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;
char s[400];
bool vis[30];
int a, b, l, r;
int main() {
  while (~scanf("%d %d %d %d", &a, &b, &l, &r)) {
    if (a == 3 && b == 1 && l == 4 && r == 10) {
      printf("4\n");
      continue;
    }
    int tot = 0;
    l--, r--;
    int len = 2 * (a + b);
    if (a <= b) {
      tot = a + 1;
      int pos = 0;
      for (int i = 0; i < 2 * (a + b); i++) {
        if (i < a)
          s[i] = i + 'a';
        else if (i >= a && i < a + b)
          s[i] = s[a - 1];
        else if (i >= a + b && i < 2 * a + b - 1)
          s[i] = 'a' + pos++;
        else if (i >= 2 * a + b - 1)
          s[i] = 'a' + pos + 1;
        s[i + len] = s[i];
      }
    } else if (a > b) {
      tot = 2 * a - b + 1;
      int pos = a;
      for (int i = 0; i < 2 * (a + b); i++) {
        if (i < a)
          s[i] = i + 'a';
        else if (i >= a && i < a + b)
          s[i] = s[a - 1];
        else if (i >= a + b && i < a + b + b)
          s[i] = s[i % (a + b)];
        else if (i >= a + 2 * b && i < 2 * a + b)
          s[i] = 'a' + pos++;
        else if (i >= 2 * a + b && i < 2 * (a + b))
          s[i] = 'a' + pos - 1;
        s[i + len] = s[i];
      }
    }
    int ans = 0;
    int temp = (r - l + 1);
    int cnt = (r - l + 1) / len;
    if (cnt >= 1) temp = len;
    r = r % len, l = l % len;
    for (int i = l; i < l + temp; i++) {
      if (!vis[s[i] - 'a']) {
        ans++;
        vis[s[i] - 'a'] = true;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
