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 i, i0, n, m, ans, a, b, l, r;
char s[100005];
set<char> S;
int cal(char x, char y, char z) {
  for (int i = 1; i <= a; i++) s[i] = 'a' + i - 1;
  int cnt = 0;
  int las = a;
  while (las <= r) {
    for (int i = las + 1; i <= las + b; i++) {
      if (cnt == 0) s[i] = x;
      if (cnt == 1) s[i] = y;
      if (cnt == 2) s[i] = z;
    }
    cnt++;
    if (cnt == 3) break;
    S.clear();
    for (int i = las + b; i >= las + b - a + 1; i--) S.insert(s[i]);
    for (int i = las + b + 1; i <= las + b + a; i++) {
      for (char c = 'a'; c <= 'z'; c++) {
        if (S.find(c) == S.end()) {
          s[i] = c;
          S.insert(c);
          break;
        }
      }
    }
    las += a + b;
  }
  S.clear();
  for (int i = l; i <= r; i++) S.insert(s[i]);
  return S.size();
}
int main() {
  scanf("%d %d %d %d", &a, &b, &l, &r);
  int d = (l - 1) / (a + b);
  l -= d * (a + b);
  r -= d * (a + b);
  if (r - l + 1 > 2 * (a + b)) r = l + 2 * (a + b);
  int ans = INT_MAX;
  for (int x = 'a'; x <= 'z'; x++)
    for (int y = 'a'; y <= 'z'; y++)
      for (int z = 'a'; z <= 'z'; z++) ans = min(ans, cal(x, y, z));
  printf("%d\n", ans);
  return 0;
}
