#include <bits/stdc++.h>
using namespace std;
int a, b, l, r;
char s[50];
bool vis[30];
int solve(int x) {
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < a; i++) s[i] = 'a' + i;
  for (int i = a; i < a + b; i++) s[i] = 'a' + x;
  for (int i = b; i < a + b; i++) vis[s[i] - 'a'] = 1;
  int pos = 0;
  for (int i = a + b; i < a * 2 + b; i++) {
    while (vis[pos]) pos++;
    s[i] = 'a' + pos;
    vis[pos] = 1;
  }
  memset(vis, 0, sizeof(vis));
  for (int i = a * 2 + b; i < 2 * (a + b); i++) s[i] = s[a * 2 + b - 1];
  int res = 0;
  if (l <= r) {
    for (int i = l; i <= r; i++)
      if (!vis[s[i] - 'a']) {
        vis[s[i] - 'a'] = 1;
        res++;
      }
  } else {
    for (int i = 0; i < 2 * (a + b); i++) {
      if (i > r && i < l) continue;
      if (!vis[s[i] - 'a']) {
        vis[s[i] - 'a'] = 1;
        res++;
      }
    }
  }
  return res;
}
int main() {
  scanf("%d%d%d%d", &a, &b, &l, &r);
  l--;
  r--;
  if (r - l + 1 >= 2 * (a + b)) {
    int ans = a + max(1, a - b);
    printf("%d\n", ans);
    return 0;
  }
  l %= 2 * (a + b);
  r %= 2 * (a + b);
  int ans = 0x7fffffff;
  for (int i = 0; i < a; i++) {
    ans = min(ans, solve(i));
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in CPP for 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>
using namespace std;
int a, b, l, r;
char s[50];
bool vis[30];
int solve(int x) {
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < a; i++) s[i] = 'a' + i;
  for (int i = a; i < a + b; i++) s[i] = 'a' + x;
  for (int i = b; i < a + b; i++) vis[s[i] - 'a'] = 1;
  int pos = 0;
  for (int i = a + b; i < a * 2 + b; i++) {
    while (vis[pos]) pos++;
    s[i] = 'a' + pos;
    vis[pos] = 1;
  }
  memset(vis, 0, sizeof(vis));
  for (int i = a * 2 + b; i < 2 * (a + b); i++) s[i] = s[a * 2 + b - 1];
  int res = 0;
  if (l <= r) {
    for (int i = l; i <= r; i++)
      if (!vis[s[i] - 'a']) {
        vis[s[i] - 'a'] = 1;
        res++;
      }
  } else {
    for (int i = 0; i < 2 * (a + b); i++) {
      if (i > r && i < l) continue;
      if (!vis[s[i] - 'a']) {
        vis[s[i] - 'a'] = 1;
        res++;
      }
    }
  }
  return res;
}
int main() {
  scanf("%d%d%d%d", &a, &b, &l, &r);
  l--;
  r--;
  if (r - l + 1 >= 2 * (a + b)) {
    int ans = a + max(1, a - b);
    printf("%d\n", ans);
    return 0;
  }
  l %= 2 * (a + b);
  r %= 2 * (a + b);
  int ans = 0x7fffffff;
  for (int i = 0; i < a; i++) {
    ans = min(ans, solve(i));
  }
  printf("%d\n", ans);
  return 0;
}

```