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;
const int maxn = 100;
int a, b, l, r;
int x[maxn], vis[30];
int ans;
int main() {
  scanf("%d%d%d%d", &a, &b, &l, &r);
  l--, r--;
  if (a <= b) {
    if (r - l + 1 >= 2 * (a + b))
      printf("%d\n", a + 1);
    else {
      for (int i = 0; i < a; i++) x[i] = i;
      for (int i = a; i < a + b; i++) x[i] = a - 1;
      for (int i = a + b; i < 2 * a + b; i++) x[i] = i - a - b;
      x[2 * a + b - 1] = a;
      for (int i = 2 * a + b; i < 2 * (a + b); i++) x[i] = a;
      l %= (2 * (a + b));
      r %= (2 * (a + b));
      if (l <= r) {
        for (int i = l; i <= r; i++)
          if (!vis[x[i]]) vis[x[i]] = 1, ans++;
      } else {
        for (int i = l; i < 2 * (a + b); i++)
          if (!vis[x[i]]) vis[x[i]] = 1, ans++;
        for (int i = 0; i <= r; i++)
          if (!vis[x[i]]) vis[x[i]] = 1, ans++;
      }
      if (a >= 3 && r == a - 2 && l >= a && l <= a + b - 1) ans--;
      printf("%d\n", ans);
    }
  } else {
    if (r - l + 1 >= 2 * (a + b))
      printf("%d\n", 2 * a - b);
    else {
      for (int i = 0; i < a; i++) x[i] = i;
      for (int i = 0; i < b; i++) x[a + i] = a - 1;
      for (int i = 0; i < b; i++) x[a + b + i] = i;
      for (int i = 0; i < a - b; i++) x[a + 2 * b + i] = a + i;
      for (int i = 0; i < b; i++) x[2 * a + b + i] = 2 * a - b - 1;
      l %= (2 * (a + b));
      r %= (2 * (a + b));
      if (l <= r) {
        for (int i = l; i <= r; i++)
          if (!vis[x[i]]) vis[x[i]] = 1, ans++;
      } else {
        for (int i = l; i < 2 * (a + b); i++)
          if (!vis[x[i]]) vis[x[i]] = 1, ans++;
        for (int i = 0; i <= r; i++)
          if (!vis[x[i]]) vis[x[i]] = 1, ans++;
      }
      if (a >= 3 && r == a - 2 && l >= a && l <= a + b - 1) ans--;
      printf("%d\n", ans);
    }
  }
  return 0;
}
