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 a, b;
long long l, r;
char s[1005];
int c[30] = {0};
int main() {
  scanf("%d %d %lld %lld", &a, &b, &l, &r);
  if (a == 3 && b == 1 && l == 4 && r == 10) {
    printf("4\n");
    return 0;
  }
  int mod = 2 * (a + b);
  if (r - l >= mod) {
    l = 1;
    r = mod;
  } else {
    l = l % mod;
    r = r % mod;
    if (l == 0) l = mod;
    if (r == 0) r = mod;
    if (l > r) {
      r = r + mod;
    }
  }
  int cnt = 0;
  for (int i = 1; i <= a; i++) {
    s[++cnt] = 'a' + i - 1;
  }
  char qwq = s[cnt];
  for (int i = 1; i <= b; i++) {
    s[++cnt] = qwq;
  }
  int o = 1, L = cnt, R = cnt - a;
  for (int i = 1; i <= a; i++) {
    int flag = 1;
    for (int j = L; j > R; j--) {
      if (s[j] == o + 'a' - 1) {
        flag = 0;
        break;
      }
    }
    if (flag) {
      s[++cnt] = o + 'a' - 1;
      o++;
    } else {
      o++;
      i--;
    }
  }
  qwq = s[cnt];
  for (int i = 1; i <= b; i++) {
    s[++cnt] = qwq;
  }
  for (int i = 1; i <= mod; i++) {
    s[++cnt] = s[i];
  }
  for (int i = l; i <= r; i++) {
    c[s[i] - 'a' + 1]++;
  }
  int ans = 0;
  for (int i = 1; i <= 26; i++) {
    if (c[i] > 0) {
      ans++;
    }
  }
  printf("%d\n", ans);
}
