Sometimes Mister B has free evenings when he doesn't know what to do. Fortunately, Mister B found a new game, where the player can play against aliens.

All characters in this game are lowercase English letters. There are two players: Mister B and his competitor.

Initially the players have a string s consisting of the first a English letters in alphabetical order (for example, if a = 5, then s equals to "abcde").

The players take turns appending letters to string s. Mister B moves first.

Mister B must append exactly b letters on each his move. He can arbitrary choose these letters. His opponent adds exactly a letters on each move.

Mister B quickly understood that his opponent was just a computer that used a simple algorithm. The computer on each turn considers the suffix of string s of length a and generates a string t of length a such that all letters in the string t are distinct and don't appear in the considered suffix. From multiple variants of t lexicographically minimal is chosen (if a = 4 and the suffix is "bfdd", the computer chooses string t equal to "aceg"). After that the chosen string t is appended to the end of s.

Mister B soon found the game boring and came up with the following question: what can be the minimum possible number of different letters in string s on the segment between positions l and r, inclusive. Letters of string s are numerated starting from 1.

Input

First and only line contains four space-separated integers: a, b, l and r (1 ≤ a, b ≤ 12, 1 ≤ l ≤ r ≤ 109) — the numbers of letters each player appends and the bounds of the segment.

Output

Print one integer — the minimum possible number of different letters in the segment from position l to position r, inclusive, in string s.

Examples

Input

1 1 1 8


Output

2

Input

4 2 2 6


Output

3

Input

3 7 4 6


Output

1

Note

In the first sample test one of optimal strategies generate string s = "abababab...", that's why answer is 2.

In the second sample test string s = "abcdbcaefg..." can be obtained, chosen segment will look like "bcdbc", that's why answer is 3.

In the third sample test string s = "abczzzacad..." can be obtained, chosen, segment will look like "zzz", that's why answer is 1.
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Pradyumn Agrawal coderbond007
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, FastReader in, PrintWriter out) {
            int a = in.nextInt();
            int b = in.nextInt();
            int l = in.nextInt() - 1;
            int r = in.nextInt() - 1;
            int ab = a + b;
            int m = 4;
            int[] cc = new int[ab * m];
            int h = l / ab;
            l -= h * ab;
            r -= h * ab;
            h = (r - l + 1) / ab;
            if (h > 2)
                r -= (h - 2) * ab;
            int d = a > b ? a - b : 1;
            for (h = 0; h < m; h++) {
                for (int k = 0; k < a - d; k++)
                    cc[(a + b) * h + k] = k;
                for (int k = 0; k < d; k++)
                    cc[(a + b) * h + a - d + k] = h % 2 == 0 ? a - d + k : a + k;
            }
            int min = a + a;
            for (int k0 = 1; k0 <= d; k0++) {
                for (int k = 0; k < b; k++)
                    cc[(a + b) * 0 + a + k] = cc[(a + b) * 0 + a - k0];
                for (int k1 = 1; k1 <= d; k1++) {
                    for (int k = 0; k < b; k++)
                        cc[(a + b) * 1 + a + k] = cc[(a + b) * 1 + a - k1];
                    for (int k2 = 1; k2 <= d; k2++) {
                        for (int k = 0; k < b; k++)
                            cc[(a + b) * 2 + a + k] = cc[(a + b) * 2 + a - k2];
                        for (int k3 = 1; k3 <= d; k3++) {
                            for (int k = 0; k < b; k++)
                                cc[(a + b) * 3 + a + k] = cc[(a + b) * 3 + a - k3];
                            min = Math.min(min, count(cc, l, r));
                        }
                    }
                }
            }
            out.println(min);
        }

        static int count(int[] cc, int l, int r) {
            boolean[] used = new boolean[24];
            int cnt = 0;
            for (int i = l; i <= r; i++) {
                int c = cc[i];
                if (!used[c]) {
                    used[c] = true;
                    cnt++;
                }
            }
            return cnt;
        }

    }

    static class FastReader {
        private InputStream stream;
        private byte[] buf = new byte[8192];
        private int curChar;
        private int pnumChars;
        private FastReader.SpaceCharFilter filter;

        public FastReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (pnumChars == -1) {
                throw new InputMismatchException();
            }
            if (curChar >= pnumChars) {
                curChar = 0;
                try {
                    pnumChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (pnumChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

        public int nextInt() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                if (c == ',') {
                    c = read();
                }
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public boolean isSpaceChar(int c) {
            if (filter != null) {
                return filter.isSpaceChar(c);
            }
            return isWhitespace(c);
        }

        public static boolean isWhitespace(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);

        }

    }
}