#include <bits/stdc++.h>
int solve() {
  int a, b, l, r, ans, lr, rr, lq, rq;
  scanf("%d%d%d%d", &a, &b, &l, &r);
  lq = (l - 1) / (a + b);
  lr = (l - 1) % (a + b) + 1;
  rq = (r - 1) / (a + b);
  rr = (r - 1) % (a + b) + 1;
  if (lq == rq) {
    if (rr <= a)
      ans = rr - lr + 1;
    else if (lr <= a)
      ans = a - lr + 1;
    else
      ans = 1;
  } else {
    rr = std::min(rr, a);
    if (b < a) {
      if (rq - lq >= 3)
        ans = 2 * a - b;
      else if (rq - lq == 1) {
        if (lr > a)
          ans = 1 + rr;
        else
          ans = std::min(a - b, a - lr + 1) + rr;
      } else if (rq - lq == 2) {
        if (lr > a)
          ans = std::max(b + 1, rr) + a - b;
        else
          ans = std::max(b + a - lr + 1, rr) + a - b;
      }
    } else {
      if (rq - lq >= 2)
        ans = a + 1;
      else {
        if (rr >= a)
          ans = a + 1;
        else if (lr > a)
          ans = 1 + rr;
        else
          ans = std::min(rr + a - lr + 1, a);
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}
int main(int argc, char* argv[]) { solve(); }

### Prompt

In cpp, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
int solve() {
  int a, b, l, r, ans, lr, rr, lq, rq;
  scanf("%d%d%d%d", &a, &b, &l, &r);
  lq = (l - 1) / (a + b);
  lr = (l - 1) % (a + b) + 1;
  rq = (r - 1) / (a + b);
  rr = (r - 1) % (a + b) + 1;
  if (lq == rq) {
    if (rr <= a)
      ans = rr - lr + 1;
    else if (lr <= a)
      ans = a - lr + 1;
    else
      ans = 1;
  } else {
    rr = std::min(rr, a);
    if (b < a) {
      if (rq - lq >= 3)
        ans = 2 * a - b;
      else if (rq - lq == 1) {
        if (lr > a)
          ans = 1 + rr;
        else
          ans = std::min(a - b, a - lr + 1) + rr;
      } else if (rq - lq == 2) {
        if (lr > a)
          ans = std::max(b + 1, rr) + a - b;
        else
          ans = std::max(b + a - lr + 1, rr) + a - b;
      }
    } else {
      if (rq - lq >= 2)
        ans = a + 1;
      else {
        if (rr >= a)
          ans = a + 1;
        else if (lr > a)
          ans = 1 + rr;
        else
          ans = std::min(rr + a - lr + 1, a);
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}
int main(int argc, char* argv[]) { solve(); }

```