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, l, r, A[1010];
bool vis[26];
inline bool check(int l, int r, int x) {
  for (int i = max(1, l); i <= r; i++)
    if (A[i] == x) return true;
  return false;
}
int solve(int s) {
  int rear = 0;
  for (int i = 0; i < 4; i++) {
    for (int j = 1, k = 0; j <= a; j++) {
      while (check(rear - a + 1, rear, k)) k++;
      A[rear + j] = k++;
    }
    int v;
    if ((s >> i) & 1) {
      int mx = -1;
      for (int j = a > b ? b + 1 : 1; j <= a; j++) {
        mx = max(mx, A[rear + j]);
      }
      v = mx;
    } else {
      int mn = 100;
      for (int j = a > b ? b + 1 : 1; j <= a; j++) {
        mn = min(mn, A[rear + j]);
      }
      v = mn;
    }
    for (int j = 1; j <= b; j++) {
      A[rear + a + j] = v;
    }
    rear += a + b;
  }
  memset(vis, 0, sizeof vis);
  int res = 0;
  for (int i = l; i <= r; i++) {
    if (!vis[A[i]]) {
      vis[A[i]] = true;
      res++;
    }
  }
  return res;
}
int main() {
  int L, R;
  scanf("%d%d%d%d", &a, &b, &L, &R);
  l = L % (a + b);
  if (l == 0) l = a + b;
  r = l + (R - L);
  if (r > 4 * (a + b)) r -= (r / (a + b) - 4) * (a + b);
  int ans = 100;
  for (int i = 0; i < 16; i++) {
    ans = min(ans, solve(i));
  }
  printf("%d\n", ans);
  return 0;
}
