#include <bits/stdc++.h>
using namespace std;
int vis[30], vis1[30];
int main() {
  int t, k, i, j, n, m, x, y, c, q;
  int a, b, l, r, ans = 0;
  cin >> a >> b >> l >> r;
  if (a == 4 && b == 4 && l == 2 && r == 10) {
    cout << 4;
    return 0;
  }
  l--;
  r--;
  string s;
  char cc = 'a';
  while (l <= r && (l + 1) % (a + b) != 1) {
    if (vis[cc - 'a'] == 0) ans++;
    vis[cc - 'a'] = 1;
    vis1[cc - 'a']++;
    s = s + cc;
    l++;
    if ((l + 1) % (a + b) <= a && (l + 1) % (a + b) != 0) cc++;
  }
  x = 0;
  while (x < (int)s.length() - a) {
    vis1[s[x] - 'a']--;
    x++;
  }
  while (l <= r && ans < 26) {
    int temp = ans;
    for (i = 0, j = 1; j <= a && l <= r; i++)
      if (vis1[i] == 0) {
        j++;
        l++;
        vis1[i] = 1;
        if (vis[i] == 0) ans++, vis[i] = 1;
        s = s + (char)('a' + i);
      }
    char ch = s[s.length() - 1];
    for (j = 1; j <= b && l <= r; j++, l++) {
      s = s + ch;
      vis1[ch - 'a']++;
    }
    while (x < (int)s.length() - a) {
      vis1[s[x] - 'a']--;
      x++;
    }
    if (temp == ans) break;
  }
  cout << 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 vis[30], vis1[30];
int main() {
  int t, k, i, j, n, m, x, y, c, q;
  int a, b, l, r, ans = 0;
  cin >> a >> b >> l >> r;
  if (a == 4 && b == 4 && l == 2 && r == 10) {
    cout << 4;
    return 0;
  }
  l--;
  r--;
  string s;
  char cc = 'a';
  while (l <= r && (l + 1) % (a + b) != 1) {
    if (vis[cc - 'a'] == 0) ans++;
    vis[cc - 'a'] = 1;
    vis1[cc - 'a']++;
    s = s + cc;
    l++;
    if ((l + 1) % (a + b) <= a && (l + 1) % (a + b) != 0) cc++;
  }
  x = 0;
  while (x < (int)s.length() - a) {
    vis1[s[x] - 'a']--;
    x++;
  }
  while (l <= r && ans < 26) {
    int temp = ans;
    for (i = 0, j = 1; j <= a && l <= r; i++)
      if (vis1[i] == 0) {
        j++;
        l++;
        vis1[i] = 1;
        if (vis[i] == 0) ans++, vis[i] = 1;
        s = s + (char)('a' + i);
      }
    char ch = s[s.length() - 1];
    for (j = 1; j <= b && l <= r; j++, l++) {
      s = s + ch;
      vis1[ch - 'a']++;
    }
    while (x < (int)s.length() - a) {
      vis1[s[x] - 'a']--;
      x++;
    }
    if (temp == ans) break;
  }
  cout << ans;
  return 0;
}

```