#include <bits/stdc++.h>
int count(int *cc, int l, int r) {
  static char used[24];
  int i, cnt;
  memset(used, 0, sizeof used);
  cnt = 0;
  for (i = l; i <= r; i++)
    if (!used[cc[i]]) {
      used[cc[i]] = 1;
      cnt++;
    }
  return cnt;
}
int main() {
  static int cc[(24 * 4)];
  int a, b, ab, d, l, r, m, k0, k1, k2, k3, k, h, min;
  scanf("%d%d%d%d", &a, &b, &l, &r);
  l--, r--;
  ab = a + b;
  m = 4;
  h = l / ab;
  r -= h * ab;
  l -= h * ab;
  h = (r - l + 1) / ab;
  if (h > 2) r -= (h - 2) * ab;
  d = a > b ? a - b : 1;
  for (h = 0; h < m; h++) {
    for (k = 0; k < a - d; k++) cc[h * ab + k] = k;
    for (k = 0; k < d; k++)
      cc[h * ab + a - d + k] = h % 2 == 0 ? a - d + k : a + k;
  }
  min = a + a;
  for (k0 = 1; k0 <= d; k0++) {
    for (k = 0; k < b; k++) cc[ab * 0 + a + k] = cc[ab * 0 + a - k0];
    for (k1 = 1; k1 <= d; k1++) {
      for (k = 0; k < b; k++) cc[ab * 1 + a + k] = cc[ab * 1 + a - k1];
      for (k2 = 1; k2 <= d; k2++) {
        for (k = 0; k < b; k++) cc[ab * 2 + a + k] = cc[ab * 2 + a - k2];
        for (k3 = 1; k3 <= d; k3++) {
          int cnt;
          for (k = 0; k < b; k++) cc[ab * 3 + a + k] = cc[ab * 3 + a - k3];
          if (min > (cnt = count(cc, l, r))) min = cnt;
        }
      }
    }
  }
  printf("%d\n", min);
  return 0;
}

### Prompt

Please create a solution in cpp to 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 count(int *cc, int l, int r) {
  static char used[24];
  int i, cnt;
  memset(used, 0, sizeof used);
  cnt = 0;
  for (i = l; i <= r; i++)
    if (!used[cc[i]]) {
      used[cc[i]] = 1;
      cnt++;
    }
  return cnt;
}
int main() {
  static int cc[(24 * 4)];
  int a, b, ab, d, l, r, m, k0, k1, k2, k3, k, h, min;
  scanf("%d%d%d%d", &a, &b, &l, &r);
  l--, r--;
  ab = a + b;
  m = 4;
  h = l / ab;
  r -= h * ab;
  l -= h * ab;
  h = (r - l + 1) / ab;
  if (h > 2) r -= (h - 2) * ab;
  d = a > b ? a - b : 1;
  for (h = 0; h < m; h++) {
    for (k = 0; k < a - d; k++) cc[h * ab + k] = k;
    for (k = 0; k < d; k++)
      cc[h * ab + a - d + k] = h % 2 == 0 ? a - d + k : a + k;
  }
  min = a + a;
  for (k0 = 1; k0 <= d; k0++) {
    for (k = 0; k < b; k++) cc[ab * 0 + a + k] = cc[ab * 0 + a - k0];
    for (k1 = 1; k1 <= d; k1++) {
      for (k = 0; k < b; k++) cc[ab * 1 + a + k] = cc[ab * 1 + a - k1];
      for (k2 = 1; k2 <= d; k2++) {
        for (k = 0; k < b; k++) cc[ab * 2 + a + k] = cc[ab * 2 + a - k2];
        for (k3 = 1; k3 <= d; k3++) {
          int cnt;
          for (k = 0; k < b; k++) cc[ab * 3 + a + k] = cc[ab * 3 + a - k3];
          if (min > (cnt = count(cc, l, r))) min = cnt;
        }
      }
    }
  }
  printf("%d\n", min);
  return 0;
}

```