#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  char s[105], c = 'a';
  for (int i = 1; i <= a; i++) {
    s[i] = c;
    c++;
  }
  bool mark[200];
  int li;
  for (int i = a + 1; i <= 98; i++) {
    if (i % (a + b) > a || i % (a + b) == 0) {
      s[i] = s[i - 1];
      li = i;
    } else {
      for (int j = 'a'; j <= 'z'; j++) mark[j] = 0;
      for (int j = li; j >= li - a + 1; j--) {
        mark[s[j]] = 1;
      }
      int k = i + a - 1;
      for (int j = i; j <= k; j++, i++) {
        for (int p = 'a'; p <= 'z'; p++) {
          if (!mark[p]) {
            mark[p] = 1;
            s[i] = p;
            break;
          }
        }
      }
      i--;
    }
  }
  s[0] = '9';
  s[100] = 0;
  int p = l % (a + b);
  if (p == 0) p = a + b;
  int d = 0;
  r -= (l - p);
  for (int i = 'a'; i <= 'z'; i++) {
    mark[i] = 0;
  }
  for (p; p <= 98 && p <= r; p++) {
    if (!mark[s[p]]) {
      mark[s[p]] = 1;
      d++;
    }
  }
  if ((r - l + 1) > (a + b) && (r - l + 1) <= (2 * a + b) &&
      (l % (a + b) > a || l % (a + b) == 0) && r % (a + b) > b &&
      r % (a + b) < a) {
    d--;
  }
  printf("%d\n", d);
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>
using namespace std;
int main() {
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  char s[105], c = 'a';
  for (int i = 1; i <= a; i++) {
    s[i] = c;
    c++;
  }
  bool mark[200];
  int li;
  for (int i = a + 1; i <= 98; i++) {
    if (i % (a + b) > a || i % (a + b) == 0) {
      s[i] = s[i - 1];
      li = i;
    } else {
      for (int j = 'a'; j <= 'z'; j++) mark[j] = 0;
      for (int j = li; j >= li - a + 1; j--) {
        mark[s[j]] = 1;
      }
      int k = i + a - 1;
      for (int j = i; j <= k; j++, i++) {
        for (int p = 'a'; p <= 'z'; p++) {
          if (!mark[p]) {
            mark[p] = 1;
            s[i] = p;
            break;
          }
        }
      }
      i--;
    }
  }
  s[0] = '9';
  s[100] = 0;
  int p = l % (a + b);
  if (p == 0) p = a + b;
  int d = 0;
  r -= (l - p);
  for (int i = 'a'; i <= 'z'; i++) {
    mark[i] = 0;
  }
  for (p; p <= 98 && p <= r; p++) {
    if (!mark[s[p]]) {
      mark[s[p]] = 1;
      d++;
    }
  }
  if ((r - l + 1) > (a + b) && (r - l + 1) <= (2 * a + b) &&
      (l % (a + b) > a || l % (a + b) == 0) && r % (a + b) > b &&
      r % (a + b) < a) {
    d--;
  }
  printf("%d\n", d);
}

```