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 num[100];
int main() {
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  for (int i = 1; i <= a; i++) {
    num[i] = i;
  }
  for (int i = 1; i <= b; i++) {
    num[a + i] = a;
  }
  int vis[50];
  memset(vis, 0, sizeof(vis));
  for (int i = a + b; i > b; i--) {
    vis[num[i]]++;
  }
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j < 50; j++) {
      if (!vis[j]) {
        vis[j]++;
        num[a + b + i] = j;
        break;
      }
    }
  }
  for (int i = a + a + b + 1; i <= a + a + b + b; i++) {
    num[i] = num[a + a + b];
  }
  int n = a + a + b + b;
  if (r - l >= n) {
    int ans = 0;
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; i++) {
      if (!vis[num[i]]) {
        vis[num[i]]++;
        ans++;
      }
    }
    cout << ans << endl;
  } else {
    int k = l / n;
    l -= k * n;
    r -= k * n;
    int ans = 0;
    int used[50], fst = 0;
    memset(vis, 0, sizeof(vis));
    for (int i = l; i <= r; i++) {
      int kk = i;
      if (kk <= n) {
        if (!vis[num[kk]]) {
          vis[num[kk]]++;
          ans++;
          used[fst++] = num[kk];
        }
      } else {
        break;
      }
    }
    for (int i = n + 1; i <= r; i++) {
      if (i <= n + a) {
        int vis2[50] = {0};
        for (int j = i - 1; j > n - a; j--) {
          vis2[num[j]]++;
        }
        int flag = 0;
        for (int j = 0; j < fst; j++) {
          if (!vis2[used[j]]) {
            flag = 1;
            num[i] = used[j];
            break;
          }
        }
        if (!flag) {
          for (int j = 1; j <= 50; j++) {
            if (!vis2[j]) {
              used[fst++] = j;
              ans++;
              num[i] = j;
              break;
            }
          }
        }
      } else if (i <= n + a + a + b && i > n + a + b) {
        int vis2[50] = {0};
        for (int j = i; j > n + b; j--) {
          vis2[num[i]]++;
        }
        int flag = 0;
        for (int j = 0; j < fst; j++) {
          if (!vis2[used[j]]) {
            flag = 1;
            num[i] = used[j];
            break;
          }
        }
        if (!flag) {
          for (int j = 1; j <= 50; j++) {
            if (!vis2[j]) {
              used[fst++] = j;
              ans++;
              num[i] = j;
              break;
            }
          }
        }
      } else {
        num[i] = num[i - 1];
      }
    }
    cout << ans << endl;
  }
  return 0;
}
