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 v[110], A[2010], ans = 10010, f[20], a, b, l, r, L, R;
inline bool Apr(int l, int r, int x) {
  for (int i = max(1, l); i <= r; i++)
    if (A[i] == x) return (1);
  return (0);
}
inline void Work() {
  int lst = 0;
  for (int i = 1; i <= 4; i++) {
    if (!f[i * 2 - 1]) {
      for (int k = 0, j = 1; j <= a; j++) {
        while (Apr(lst - a + 1, lst, k)) k++;
        A[lst + j] = k, k++;
      }
    } else {
      for (int k = 0, j = a; j >= 1; j--) {
        while (Apr(lst - a + 1, lst, k)) k++;
        A[lst + j] = k, k++;
      }
    }
    if (!f[i * 2]) {
      int mn = 100;
      for (int j = 1; j <= a; j++) mn = min(mn, A[lst + j]);
      for (int j = 1; j <= b; j++) A[lst + a + j] = mn;
    } else {
      int mx = 0;
      for (int j = 1; j <= a; j++) mx = max(mx, A[lst + j]);
      for (int j = 1; j <= b; j++) A[lst + a + j] = mx;
    }
    lst += a + b;
  }
  int nm = 0;
  memset(v, 0, sizeof(v));
  for (int i = l; i <= r; i++) v[A[i]] = 1;
  for (int i = 0; i <= 100; i++) nm += v[i];
  ans = min(ans, nm);
}
inline void Dfs(int x) {
  if (x > 10) {
    Work();
    return;
  }
  f[x] = 0;
  Dfs(x + 1);
  f[x] = 1;
  Dfs(x + 1);
  f[x] = 0;
}
int main() {
  scanf("%d%d%d%d", &a, &b, &L, &R);
  l = L % (a + b);
  if (!l) l = a + b;
  r = l + (R - L);
  if (r > (a + b) * 4) r -= (r / (a + b) - 4) * (a + b);
  Dfs(2);
  printf("%d\n", ans);
}
