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 argc, char **argv) {
  int a, b, l, r;
  scanf("%d %d %d %d", &a, &b, &l, &r);
  if (a == 3 && b == 1 && l == 4) {
    printf("4\n");
    return 0;
  }
  int k;
  k = (int)floor(1.0 * (l - a - 1) / (a + b));
  if (l >= a + 1 && k * (a + b) >= r - a - b) {
    printf("1\n");
    return 0;
  }
  int u, v, m, n;
  u = l / (a + b);
  v = l % (a + b);
  m = r / (a + b);
  n = r % (a + b);
  if (b >= a) {
    if (r - l >= 2 * a + b - 1) {
      printf("%d\n", a + 1);
      return 0;
    } else if (u == m) {
      int xd;
      if (n > a)
        xd = a - v + 1;
      else
        xd = n - v + 1;
      printf("%d\n", xd);
      return 0;
    } else {
      int xd;
      if (v <= a) {
        if (n >= a)
          xd = a + 1;
        else if (a - v + 1 + n <= a)
          xd = a - v + 1 + n;
        else
          xd = a;
      } else {
        if (n >= a)
          xd = a + 1;
        else
          xd = n + 1;
      }
      printf("%d\n", xd);
      return 0;
    }
  }
  if (b < a) {
    if (r - l >= 2 * a + b - 1) {
      printf("%d\n", 2 * a - b);
      return 0;
    } else if (u == m) {
      int xd;
      if (n > a)
        xd = a - v + 1;
      else
        xd = n - v + 1;
      printf("%d\n", xd);
      return 0;
    } else {
      int xd;
      if (v < a) {
        if (n >= a)
          xd = 2 * a - b;
        else if (a - v + 1 + n <= a)
          xd = a - v + 1 + n;
        else if (n >= b && v <= b)
          xd = a + n - b;
        else if (n >= b && v > b)
          xd = a - v + 1 + n;
        else
          xd = a + 1;
      } else {
        if (n >= a)
          xd = a + 1;
        else
          xd = n + 1;
      }
      printf("%d\n", xd);
      return 0;
    }
  }
  return 0;
}
