#include <bits/stdc++.h>
using namespace std;
int a, b, l, r;
int tool[26];
char K[100];
int F(int L, int R) {
  memset(tool, 0, sizeof(tool));
  int cnt = 0;
  for (int i = L; i <= R; i++) {
    if (tool[K[i] - 'a'] == 0) {
      tool[K[i] - 'a'] = 1;
      cnt++;
    }
  }
  return cnt;
}
int main() {
  cin >> a >> b >> l >> r;
  l--;
  r--;
  if (r - l + 1 >= (a + b) * 2) {
    l = 0;
    r = (a + b) * 2 - 1;
  } else {
    l %= (a + b) * 2;
    r %= (a + b) * 2;
    if (r < l) r += (a + b) * 2;
  }
  int i;
  memset(tool, 0, sizeof(tool));
  int ans;
  if (a > b + 1) {
    for (i = 0; i < b; i++) {
      K[i + a + b] = K[i] = 'a' + i;
    }
    for (i = b; i < a; i++) {
      K[i] = 'a' + i;
      K[i + a + b] = K[i] + a - b;
    }
    for (i = a; i < a + b; i++) {
      K[i] = K[b + 1];
      K[i + a + b] = K[a + b * 2 + 1];
    }
    for (i = 0; i < a * 2 + b * 2; i++) {
      K[i + a * 2 + b * 2] = K[i];
    }
    ans = F(l, r);
    for (i = a; i < a + b; i++) {
      K[i] = K[a - 2];
      K[i + a + b] = K[a * 2 + b - 2];
    }
    for (i = 0; i < a * 2 + b * 2; i++) {
      K[i + a * 2 + b * 2] = K[i];
    }
    ans = min(ans, F(l, r));
  } else {
    for (i = 0; i < a; i++) {
      K[i] = K[i + a + b] = 'a' + i;
    }
    for (i = a; i < a + b; i++) K[i] = K[i - 1];
    K[a * 2 + b - 1]++;
    for (i = a + a + b; i < (a + b) * 2; i++) K[i] = K[i - 1];
    for (i = 0; i < a * 2 + b * 2; i++) {
      K[i + a * 2 + b * 2] = K[i];
    }
    ans = F(l, r);
  }
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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>
using namespace std;
int a, b, l, r;
int tool[26];
char K[100];
int F(int L, int R) {
  memset(tool, 0, sizeof(tool));
  int cnt = 0;
  for (int i = L; i <= R; i++) {
    if (tool[K[i] - 'a'] == 0) {
      tool[K[i] - 'a'] = 1;
      cnt++;
    }
  }
  return cnt;
}
int main() {
  cin >> a >> b >> l >> r;
  l--;
  r--;
  if (r - l + 1 >= (a + b) * 2) {
    l = 0;
    r = (a + b) * 2 - 1;
  } else {
    l %= (a + b) * 2;
    r %= (a + b) * 2;
    if (r < l) r += (a + b) * 2;
  }
  int i;
  memset(tool, 0, sizeof(tool));
  int ans;
  if (a > b + 1) {
    for (i = 0; i < b; i++) {
      K[i + a + b] = K[i] = 'a' + i;
    }
    for (i = b; i < a; i++) {
      K[i] = 'a' + i;
      K[i + a + b] = K[i] + a - b;
    }
    for (i = a; i < a + b; i++) {
      K[i] = K[b + 1];
      K[i + a + b] = K[a + b * 2 + 1];
    }
    for (i = 0; i < a * 2 + b * 2; i++) {
      K[i + a * 2 + b * 2] = K[i];
    }
    ans = F(l, r);
    for (i = a; i < a + b; i++) {
      K[i] = K[a - 2];
      K[i + a + b] = K[a * 2 + b - 2];
    }
    for (i = 0; i < a * 2 + b * 2; i++) {
      K[i + a * 2 + b * 2] = K[i];
    }
    ans = min(ans, F(l, r));
  } else {
    for (i = 0; i < a; i++) {
      K[i] = K[i + a + b] = 'a' + i;
    }
    for (i = a; i < a + b; i++) K[i] = K[i - 1];
    K[a * 2 + b - 1]++;
    for (i = a + a + b; i < (a + b) * 2; i++) K[i] = K[i - 1];
    for (i = 0; i < a * 2 + b * 2; i++) {
      K[i + a * 2 + b * 2] = K[i];
    }
    ans = F(l, r);
  }
  cout << ans;
  return 0;
}

```