#include <bits/stdc++.h>
using namespace std;
int main() {
  long long a, b, l, r, c, ans;
  cin >> a >> b >> l >> r;
  c = l % (a + b);
  if (!c) c = a + b;
  if (c > a) {
    ans = 1;
    l += (a + b - c + 1);
    if (l <= r) ans += min(a, r - l + 1);
    l += a;
    if (a > b) {
      l += (2 * b);
      if (l <= r) {
        ans += min(a - b, r - l + 1);
        ans--;
      }
    }
  } else {
    ans = min(a - c + 1, r - l + 1);
    l += (a - c + 1);
    if (l <= r) {
      ans++;
      l += b;
      if ((a <= b && a > r - l + 1) || a > b) ans--;
    }
    if (l <= r) ans += min(c - 1, r - l + 1);
    l += (c - 1);
    if (a > b) {
      if ((a - c + 1) > (a - b)) {
        l += (b - c + 1);
        if (l <= r) ans += min(a - b, r - l + 1);
      } else {
        if (l <= r) ans += min(a - c + 1, r - l + 1);
        l += (a - c + 1);
        l += (b + b);
        if (l <= r) ans += min(c - b - 1, r - l + 1);
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### 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() {
  long long a, b, l, r, c, ans;
  cin >> a >> b >> l >> r;
  c = l % (a + b);
  if (!c) c = a + b;
  if (c > a) {
    ans = 1;
    l += (a + b - c + 1);
    if (l <= r) ans += min(a, r - l + 1);
    l += a;
    if (a > b) {
      l += (2 * b);
      if (l <= r) {
        ans += min(a - b, r - l + 1);
        ans--;
      }
    }
  } else {
    ans = min(a - c + 1, r - l + 1);
    l += (a - c + 1);
    if (l <= r) {
      ans++;
      l += b;
      if ((a <= b && a > r - l + 1) || a > b) ans--;
    }
    if (l <= r) ans += min(c - 1, r - l + 1);
    l += (c - 1);
    if (a > b) {
      if ((a - c + 1) > (a - b)) {
        l += (b - c + 1);
        if (l <= r) ans += min(a - b, r - l + 1);
      } else {
        if (l <= r) ans += min(a - c + 1, r - l + 1);
        l += (a - c + 1);
        l += (b + b);
        if (l <= r) ans += min(c - b - 1, r - l + 1);
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```