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;
long long int h[1000];
int main() {
  long long int a, b, l, r;
  cin >> a >> b >> l >> r;
  string s[1000];
  long long int x = 0;
  if (a <= b) {
    x = 0;
    for (long long int i = 0; i < a; i++) {
      s[i] = 97 + i;
      x = i;
    }
    x++;
    for (long long int i = 0; i < b; i++) {
      s[x] = s[a - 1];
      x++;
    }
    for (long long int i = 0; i < a - 1; i++) {
      s[x] = s[i];
      x++;
    }
    s[x] = 97 + a;
    x++;
    for (long long int i = 0; i < b; i++) {
      s[x] = 97 + a;
      x++;
    }
  } else if (b < a) {
    x = 0;
    for (long long int i = 0; i < a; i++) {
      s[x] = 97 + i;
      x++;
    }
    for (long long int i = 0; i < b; i++) {
      s[x] = 97 + a - 1;
      x++;
    }
    for (long long int i = 0; i < b; i++) {
      s[x] = s[i];
      x++;
    }
    long long int y = 0;
    for (long long int i = b; i < a; i++) {
      s[x] = 97 + a + y;
      y++;
      x++;
    }
    for (long long int i = 0; i < b; i++) {
      s[x] = 97 + a;
      x++;
    }
  }
  long long int flag = 0;
  if (r - l + 1 < x && (l - 1) % (2 * (a + b)) <= (r - 1) % (2 * (a + b))) {
    for (long long int i = (l - 1) % (2 * (a + b));
         i <= (r - 1) % (2 * (a + b)); i++) {
      h[s[i][0]]++;
    }
  } else {
    if (a >= b) {
      if ((l - 1) % (2 * (a + b)) >= a - 1 &&
          (l - 1) % (2 * (a + b)) <= a + b - 1 &&
          (r - 1) % (2 * (a + b)) == a - 2) {
        flag = 1;
      }
    }
    if ((r - l + 1) >= x - 1) {
      for (long long int i = 0; i < x; i++) {
        h[s[i][0]]++;
      }
    }
    if (l == 0) {
      for (long long int i = 0; i < x; i++) {
        h[s[i][0]]++;
      }
    } else {
      for (long long int i = (l - 1) % (2 * (a + b)); i < x; i++) {
        h[s[i][0]]++;
      }
    }
    for (long long int i = 0; i <= (r - 1) % (2 * (a + b)); i++) {
      h[s[i][0]]++;
    }
  }
  long long int ans = 0;
  for (long long int i = 97; i <= 122; i++) {
    if (h[i] > 0) ans++;
  }
  if (flag == 1)
    cout << ans - 1;
  else
    cout << ans;
  return 0;
}
