An expedition group flew from planet ACM-1 to Earth in order to study the bipedal species (its representatives don't even have antennas on their heads!).

The flying saucer, on which the brave pioneers set off, consists of three sections. These sections are connected by a chain: the 1-st section is adjacent only to the 2-nd one, the 2-nd one — to the 1-st and the 3-rd ones, the 3-rd one — only to the 2-nd one. The transitions are possible only between the adjacent sections.

The spacecraft team consists of n aliens. Each of them is given a rank — an integer from 1 to n. The ranks of all astronauts are distinct. The rules established on the Saucer, state that an alien may move from section a to section b only if it is senior in rank to all aliens who are in the segments a and b (besides, the segments a and b are of course required to be adjacent). Any alien requires exactly 1 minute to make a move. Besides, safety regulations require that no more than one alien moved at the same minute along the ship.

Alien A is senior in rank to alien B, if the number indicating rank A, is more than the corresponding number for B.

At the moment the whole saucer team is in the 3-rd segment. They all need to move to the 1-st segment. One member of the crew, the alien with the identification number CFR-140, decided to calculate the minimum time (in minutes) they will need to perform this task.

Help CFR-140, figure out the minimum time (in minutes) that all the astronauts will need to move from the 3-rd segment to the 1-st one. Since this number can be rather large, count it modulo m.

Input

The first line contains two space-separated integers: n and m (1 ≤ n, m ≤ 109) — the number of aliens on the saucer and the number, modulo which you should print the answer, correspondingly.

Output

Print a single number — the answer to the problem modulo m.

Examples

Input

1 10


Output

2


Input

3 8


Output

2

Note

In the first sample the only crew member moves from segment 3 to segment 2, and then from segment 2 to segment 1 without any problems. Thus, the whole moving will take two minutes.

To briefly describe the movements in the second sample we will use value <image>, which would correspond to an alien with rank i moving from the segment in which it is at the moment, to the segment number j. Using these values, we will describe the movements between the segments in the second sample: <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>; In total: the aliens need 26 moves. The remainder after dividing 26 by 8 equals 2, so the answer to this test is 2.
//package codeforces;

import java.io.*;
import java.util.Comparator;
import java.util.Objects;
import java.util.StringTokenizer;
import java.util.function.BiFunction;
import java.util.function.Function;

public class SearchingGraph {

    static BiFunction<Integer, Integer, Integer> ADD = (x, y) -> (x + y);

    static Function<Pair<Integer, Integer>, Integer> GET_FIRST = (x) -> (x.first);
    static Function<Pair<Integer, Integer>, Integer> GET_SECOND = (x) -> (x.second);
    static Comparator<Pair<Integer, Integer>> C = Comparator.comparing(GET_FIRST).thenComparing(GET_SECOND);

    public static void main(String[] args) throws Exception {
//        long startTime = System.nanoTime();
        int t = 1;
        while (t-- > 0) {
            solve();
        }
//        long endTime = System.nanoTime();
//        err.println("Execution Time : +" + (endTime - startTime) / 1000000 + " ms");
        exit(0);
    }

    static void solve() {

        int n = in.nextInt();
        int m = in.nextInt();

        long ans = binpow(3l, (long) n, (long)m) - 1;
        if(ans < 0) ans += m;

        System.out.println(ans);
    }

    static long binpow(long a, long b, long p) {
        if (b == 0)
            return 1;
        long res = binpow(a, b / 2, p);
        if (b % 2 == 1)
            return (((res * res) % p ) * a) % p;
        else
            return (res * res) % p;
    }

    static void debug(Object... args) {
        for (Object a : args) {
            out.println(a);
        }
    }

    static void y() {
        out.println("YES");
    }

    static void n() {
        out.println("NO");
    }

    static void fail() {
        out.println("-1");
        exit(0);
    }

    static class Pair<T, R> {
        public T first;
        public R second;

        public Pair(T first, R second) {
            this.first = first;
            this.second = second;
        }

        @Override
        public boolean equals(final Object o) {
            if (this == o) {
                return true;
            }
            if (o == null || getClass() != o.getClass()) {
                return false;
            }
            final Pair<?, ?> pair = (Pair<?, ?>) o;
            return Objects.equals(first, pair.first) && Objects.equals(second, pair.second);
        }

        @Override
        public int hashCode() {
            return Objects.hash(first, second);
        }

        @Override
        public String toString() {
            return "Pair{" + "a=" + first + ", b=" + second + '}';
        }

        public T getFirst() {
            return first;
        }

        public R getSecond() {
            return second;
        }
    }

    static <T, R> Pair<T, R> make_pair(T a, R b) {
        return new Pair<>(a, b);
    }

    static class ArrayUtils {

        static int[] reverse(int[] data) {
            int[] p = new int[data.length];
            for (int i = 0, j = data.length - 1; i < data.length; i++, j--) {
                p[i] = data[j];
            }
            return p;
        }

        static void prefixSum(long[] data) {
            for (int i = 1; i < data.length; i++) {
                data[i] += data[i - 1];
            }
        }

        static void prefixSum(int[] data) {
            for (int i = 1; i < data.length; i++) {
                data[i] += data[i - 1];
            }
        }

        static long[] reverse(long[] data) {
            long[] p = new long[data.length];
            for (int i = 0, j = data.length - 1; i < data.length; i++, j--) {
                p[i] = data[j];
            }
            return p;
        }

        static char[] reverse(char[] data) {
            char[] p = new char[data.length];
            for (int i = 0, j = data.length - 1; i < data.length; i++, j--) {
                p[i] = data[j];
            }
            return p;
        }

        static int[] MergeSort(int[] A) {
            if (A.length > 1) {
                int q = A.length / 2;
                int[] left = new int[q];
                int[] right = new int[A.length - q];
                System.arraycopy(A, 0, left, 0, q);
                System.arraycopy(A, q, right, 0, A.length - q);
                int[] left_sorted = MergeSort(left);
                int[] right_sorted = MergeSort(right);
                return Merge(left_sorted, right_sorted);
            } else {
                return A;
            }
        }

        static int[] Merge(int[] left, int[] right) {
            int[] A = new int[left.length + right.length];
            int i = 0;
            int j = 0;
            for (int k = 0; k < A.length; k++) {
                // To handle left becoming empty
                if (i == left.length && j < right.length) {
                    A[k] = right[j];
                    j++;
                    continue;
                }
                // To handle right becoming empty
                if (j == right.length && i < left.length) {
                    A[k] = left[i];
                    i++;
                    continue;
                }
                if (left[i] <= right[j]) {
                    A[k] = left[i];
                    i++;
                } else {
                    A[k] = right[j];
                    j++;
                }
            }
            return A;
        }

        static long[] MergeSort(long[] A) {
            if (A.length > 1) {
                int q = A.length / 2;
                long[] left = new long[q];
                long[] right = new long[A.length - q];
                System.arraycopy(A, 0, left, 0, q);
                System.arraycopy(A, q, right, 0, A.length - q);
                long[] left_sorted = MergeSort(left);
                long[] right_sorted = MergeSort(right);
                return Merge(left_sorted, right_sorted);
            } else {
                return A;
            }
        }

        static long[] Merge(long[] left, long[] right) {
            long[] A = new long[left.length + right.length];
            int i = 0;
            int j = 0;
            for (int k = 0; k < A.length; k++) {
                // To handle left becoming empty
                if (i == left.length && j < right.length) {
                    A[k] = right[j];
                    j++;
                    continue;
                }
                // To handle right becoming empty
                if (j == right.length && i < left.length) {
                    A[k] = left[i];
                    i++;
                    continue;
                }
                if (left[i] <= right[j]) {
                    A[k] = left[i];
                    i++;
                } else {
                    A[k] = right[j];
                    j++;
                }
            }
            return A;
        }

        static int nCrModp(int n, int r, int p) {
            if (r > n - r)
                r = n - r;

            // The array C is going to store last
            // row of pascal triangle at the end.
            // And last entry of last row is nCr
            int C[] = new int[r + 1];

            C[0] = 1; // Top row of Pascal Triangle

            // One by constructs remaining rows of Pascal
            // Triangle from top to bottom
            for (int i = 1; i <= n; i++) {

                // Fill entries of current row using previous
                // row values
                for (int j = Math.min(i, r); j > 0; j--)

                    // nCj = (n-1)Cj + (n-1)C(j-1);
                    C[j] = (C[j] + C[j - 1]) % p;
            }
            return C[r];
        }

        static long nCr(int n, int r) {
            if (r > n - r)
                r = n - r;

            // The array C is going to store last
            // row of pascal triangle at the end.
            // And last entry of last row is nCr
            long C[] = new long[r + 1];

            C[0] = 1; // Top row of Pascal Triangle

            // One by constructs remaining rows of Pascal
            // Triangle from top to bottom
            for (int i = 1; i <= n; i++) {

                // Fill entries of current row using previous
                // row values
                for (int j = Math.min(i, r); j > 0; j--)

                    // nCj = (n-1)Cj + (n-1)C(j-1);
                    C[j] = (C[j] + C[j - 1]);
            }
            return C[r];
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public int[] readAllInts(int n) {
            int[] p = new int[n];
            for (int i = 0; i < n; i++) {
                p[i] = in.nextInt();
            }
            return p;
        }

        public int[] readAllInts(int n, int s) {
            int[] p = new int[n + s];
            for (int i = s; i < n + s; i++) {
                p[i] = in.nextInt();
            }
            return p;
        }

        public long[] readAllLongs(int n) {
            long[] p = new long[n];
            for (int i = 0; i < n; i++) {
                p[i] = in.nextLong();
            }
            return p;
        }

        public long[] readAllLongs(int n, int s) {
            long[] p = new long[n + s];
            for (int i = s; i < n + s; i++) {
                p[i] = in.nextLong();
            }
            return p;
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }

    static void exit(int a) {
        out.close();
        err.close();
        System.exit(a);
    }

    static InputStream inputStream = System.in;
    static OutputStream outputStream = System.out;
    static OutputStream errStream = System.err;
    static InputReader in = new InputReader(inputStream);
    static PrintWriter out = new PrintWriter(outputStream);
    static PrintWriter err = new PrintWriter(errStream);
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

}