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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  long a, b, l, r;
  cin >> a >> b >> l >> r;
  if (r - l + 1 > 2 * (a + b)) {
    int x = 2 * a - b;
    if (x < a + 1) {
      x = a + 1;
    }
    cout << x;
  } else {
    l = l % (2 * (a + b));
    if (l == 0) {
      l = 2 * (a + b);
    }
    r = r % (2 * (a + b));
    if (r == 0) {
      r = 2 * (a + b);
    }
    if (r < l) {
      r += 2 * (a + b);
    }
    int arr[4 * (a + b) + 1];
    for (int i = 1; i <= a; i++) {
      arr[i] = i;
    }
    for (int i = a + 1; i <= a + b; i++) {
      arr[i] = arr[i - 1];
    }
    int tmp = 1, k = b;
    for (int i = a + b + 1; i <= a + b + a; i++) {
      while (true) {
        int kt = 0;
        for (int j = b + 1; j <= a + b; j++) {
          if (arr[j] == tmp) {
            kt = 1;
            break;
          }
        }
        if (kt == 0) {
          break;
        } else {
          tmp++;
        }
      }
      arr[i] = tmp;
      tmp++;
    }
    for (int i = a + a + b + 1; i <= 2 * (a + b); i++) {
      arr[i] = arr[i - 1];
    }
    for (int i = 2 * (a + b) + 1; i <= 2 * (a + b) + a; i++) {
      for (int j = (2 * (a + b)); j >= 1; j--) {
        int tmp = arr[j];
        int kt = 0;
        for (int jj = 2 * (a + b) - a + 1; jj <= 2 * (a + b); jj++) {
          if (arr[jj] == tmp) {
            kt = 1;
            break;
          }
        }
        for (int jj = 2 * (a + b); jj < i; jj++) {
          if (arr[jj] == tmp) {
            kt = 1;
            break;
          }
        }
        if (kt == 0) {
          arr[i] = tmp;
          break;
        }
      }
    }
    for (int i = 2 * (a + b) + a + 1; i <= 3 * (a + b); i++) {
      arr[i] = arr[i - 1];
    }
    for (int i = 3 * (a + b) + 1; i <= 3 * (a + b) + a; i++) {
      for (int j = (3 * (a + b)); j >= 1; j--) {
        int tmp = arr[j];
        int kt = 0;
        for (int jj = 3 * (a + b) - a + 1; jj <= 3 * (a + b); jj++) {
          if (arr[jj] == tmp) {
            kt = 1;
            break;
          }
        }
        for (int jj = 3 * (a + b) + 1; jj < i; jj++) {
          if (arr[jj] == tmp) {
            kt = 1;
            break;
          }
        }
        if (kt == 0) {
          arr[i] = tmp;
          break;
        }
      }
    }
    for (int i = 3 * (a + b) + a + 1; i <= 4 * (a + b); i++) {
      arr[i] = arr[i - 1];
    }
    int result = 0;
    for (int i = 1; i <= 4 * (a + b); i++) {
    }
    for (int i = l; i <= r; i++) {
      int kt = 0;
      for (int j = l; j < i; j++) {
        if (arr[j] == arr[i]) {
          kt = 1;
          break;
        }
      }
      if (kt == 0) {
        result++;
      }
    }
    cout << result;
  }
  return 0;
}
