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>
int main() {
  int a, b, l, r;
  scanf("%d%d%d%d", &a, &b, &l, &r);
  int len = r - l + 1;
  if (len > a + b + a) {
    if (a > b) {
      printf("%d", a + a - b);
    } else {
      printf("%d", a + 1);
    }
    return 0;
  }
  int base = a + b;
  l = l - (l - 1) / base * base;
  r = l + len - 1;
  int ans = 0;
  if (l <= a && r <= a) {
    ans = len;
  } else if (l <= a && r > a && r <= base) {
    ans = a - l + 1;
  } else if (l <= a && r > base) {
    if (a > b) {
      if (r - base >= l) {
        if (l - 1 > b) {
          ans = a + r - base - l + 1;
        } else {
          ans = a + r - base - b;
        }
      } else {
        ans = r - base + a - l + 1;
      }
    } else {
      if (r - base >= l) {
        ans = a;
      } else {
        ans = r - base + a - l + 1;
      }
    }
  } else if (l > a && l <= base && r <= base) {
    ans = 1;
  } else if (l > a && l <= base && r > base && r <= base + a) {
    ans = 1 + r - base;
  } else if (l > a && l <= base && r > base + a && r <= 2 * base) {
    ans = 1 + a;
  } else if (l > a && l <= base && r > 2 * base) {
    if (a > b) {
      if (r - 2 * base > b) {
        ans = a + r - 2 * base - b;
      } else {
        ans = a + 1;
      }
    } else {
      ans = a + 1;
    }
  }
  printf("%d\n", ans);
}
