Walking along a riverside, Mino silently takes a note of something.

"Time," Mino thinks aloud.

"What?"

"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."

"And what are you recording?"

"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.

Doubtfully, Kanno peeks at Mino's records. 

The records are expressed as a string s of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).

You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer p is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.

In this problem, a positive integer p is considered a period of string s, if for all 1 ≤ i ≤ \lvert s \rvert - p, the i-th and (i + p)-th characters of s are the same. Here \lvert s \rvert is the length of s.

Input

The first line contains two space-separated integers n and p (1 ≤ p ≤ n ≤ 2000) — the length of the given string and the supposed period, respectively.

The second line contains a string s of n characters — Mino's records. s only contains characters '0', '1' and '.', and contains at least one '.' character.

Output

Output one line — if it's possible that p is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).

Examples

Input

10 7
1.0.1.0.1.


Output

1000100010


Input

10 6
1.0.1.1000


Output

1001101000


Input

10 9
1........1


Output

No

Note

In the first example, 7 is not a period of the resulting string because the 1-st and 8-th characters of it are different.

In the second example, 6 is not a period of the resulting string because the 4-th and 10-th characters of it are different.

In the third example, 9 is always a period because the only constraint that the first and last characters are the same is already satisfied.

Note that there are multiple acceptable answers for the first two examples, you can print any of them.

/**
 * Date: 12 Jun, 2018
 * Link:
 *
 * @author Prasad-Chaudhari
 * @email prasadc8897@gmail.com
 */
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Random;

public class newProgram1 {

    public static void main(String[] args) throws IOException {
        // TODO code application logic here
        FastIO in = new FastIO();
        int n = in.ni();
        int p = in.ni();
        char c[] = in.next().toCharArray();
        int a[][] = new int[p][3];
        for (int i = 0; i < n; i++) {
            if (c[i] == '.') {
                a[i % p][0]++;
            }
            if (c[i] == '0') {
                a[i % p][1]++;
            }
            if (c[i] == '1') {
                a[i % p][2]++;
            }
        }
        if (n == p) {
            System.out.println("No");
            return;
        }
        Random r = new Random();
        boolean not_period = false;
        for (int i = 0; i < p; i++) {
            if (a[i][1] != 0 && a[i][2] != 0) {
                not_period = true;
            } else {
                if (a[i][0] == 0) {
                    if (a[i][1] == 0 && a[i][2] == 0) {

                    }
                    if (a[i][1] != 0 && a[i][2] == 0) {

                    }
                    if (a[i][1] == 0 && a[i][2] != 0) {

                    }
                } else {
                    if (a[i][0] != 1) {
                        not_period = true;
                    } else {
                        if (a[i][1] != 0 || a[i][2] != 0) {
                            not_period = true;
                        }
                    }
                }
            }
        }
        if (not_period) {
            for (int i = 0; i < n; i++) {
                if (c[i] == '.') {
                    if (a[i % p][0] == 1) {
                        if (a[i % p][1] == 0) {
                            System.out.print("0");
                        } else {
                            System.out.print("1");
                        }
                    } else {
                        System.out.print("0");
                        c[i] = 0;
                        for (int j = i; j < n; j += p) {
                            if (c[j] == '.') {
                                c[j] = '1';
                                break;
                            }
                        }
                    }
                } else {
                    System.out.print(c[i]);
                }
            }
        } else {
            System.out.println("No");
        }
    }

    static class FastIO {

        private final BufferedReader br;
        private final BufferedWriter bw;
        private String s[];
        private int index;

        public FastIO() throws IOException {
            br = new BufferedReader(new InputStreamReader(System.in));
            bw = new BufferedWriter(new OutputStreamWriter(System.out, "UTF-8"));
            s = br.readLine().split(" ");
            index = 0;
        }

        public int ni() throws IOException {
            return Integer.parseInt(nextToken());
        }

        public double nd() throws IOException {
            return Double.parseDouble(nextToken());
        }

        public long nl() throws IOException {
            return Long.parseLong(nextToken());
        }

        public String next() throws IOException {
            return nextToken();
        }

        public String nli() throws IOException {
            try {
                return br.readLine();
            } catch (IOException ex) {

            }
            return null;
        }

        public int[] gia(int n) throws IOException {
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public int[] gia(int n, int start, int end) throws IOException {
            validate(n, start, end);
            int a[] = new int[n];
            for (int i = start; i < end; i++) {
                a[i] = ni();
            }
            return a;
        }

        public double[] gda(int n) throws IOException {
            double a[] = new double[n];
            for (int i = 0; i < n; i++) {
                a[i] = nd();
            }
            return a;
        }

        public double[] gda(int n, int start, int end) throws IOException {
            validate(n, start, end);
            double a[] = new double[n];
            for (int i = start; i < end; i++) {
                a[i] = nd();
            }
            return a;
        }

        public long[] gla(int n) throws IOException {
            long a[] = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public long[] gla(int n, int start, int end) throws IOException {
            validate(n, start, end);
            long a[] = new long[n];
            for (int i = start; i < end; i++) {
                a[i] = ni();
            }
            return a;
        }

        public int[][] gg(int n, int m) throws IOException {
            int adja[][] = new int[n + 1][];
            int from[] = new int[m];
            int to[] = new int[m];
            int count[] = new int[n + 1];
            for (int i = 0; i < m; i++) {
                from[i] = ni();
                to[i] = ni();
                count[from[i]]++;
                count[to[i]]++;
            }
            for (int i = 0; i <= n; i++) {
                adja[i] = new int[count[i]];
            }
            for (int i = 0; i < m; i++) {
                adja[from[i]][--count[from[i]]] = to[i];
                adja[to[i]][--count[to[i]]] = from[i];
            }
            return adja;
        }

        public void print(String s) throws IOException {
            bw.write(s);
            bw.flush();
        }

        public void println(String s) throws IOException {
            bw.write(s);
            bw.newLine();
            bw.flush();
        }

        private String nextToken() throws IndexOutOfBoundsException, IOException {
            if (index == s.length) {
                s = br.readLine().split(" ");
                index = 0;
            }
            return s[index++];
        }

        private void validate(int n, int start, int end) {
            if (start < 0 || end >= n) {
                throw new IllegalArgumentException();
            }
        }
    }
}
