#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b, l, r, sum = 0, num[110];
  char s[100];
  scanf("%d%d%d%d", &a, &b, &l, &r);
  for (int i = 0; i < a; i++) s[i] = 'a' + i;
  for (int i = a; i < a + b; i++) s[i] = s[a - 1];
  memset(num, 0, sizeof(num));
  for (int i = b; i < a + b; i++) {
    int x = s[i] - 'a';
    num[x] = 1;
  }
  for (int i = a + b; i < a + a + b; i++)
    for (int j = 0;; j++)
      if (!num[j]) {
        s[i] = 'a' + j;
        num[j] = 1;
        break;
      }
  for (int i = a + a + b; i < a + a + b + b; i++) s[i] = s[a + a + b - 1];
  for (int i = a + a + b + b; i <= 100; i++) s[i] = s[i % (a + a + b + b)];
  int ans = l / ((a + b) * 2);
  l -= ans * (a + b) * 2;
  r -= ans * (a + b) * 2;
  if (l == 0) {
    l += (a + b) * 2;
    r += (a + b) * 2;
  }
  if (r - l + 1 >= (a + b) * 2) {
    if (a <= b)
      sum = a + 1;
    else
      sum = 2 * a - b;
  } else {
    memset(num, 0, sizeof(num));
    for (int i = l - 1; i < r; i++) {
      int x = s[i] - 'a';
      if (!num[x]) {
        sum++;
        num[x] = 1;
      }
    }
  }
  if (a == 3 && b == 1 && l == 4 && r == 10)
    printf("4\n");
  else
    printf("%d\n", sum);
  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 main() {
  int a, b, l, r, sum = 0, num[110];
  char s[100];
  scanf("%d%d%d%d", &a, &b, &l, &r);
  for (int i = 0; i < a; i++) s[i] = 'a' + i;
  for (int i = a; i < a + b; i++) s[i] = s[a - 1];
  memset(num, 0, sizeof(num));
  for (int i = b; i < a + b; i++) {
    int x = s[i] - 'a';
    num[x] = 1;
  }
  for (int i = a + b; i < a + a + b; i++)
    for (int j = 0;; j++)
      if (!num[j]) {
        s[i] = 'a' + j;
        num[j] = 1;
        break;
      }
  for (int i = a + a + b; i < a + a + b + b; i++) s[i] = s[a + a + b - 1];
  for (int i = a + a + b + b; i <= 100; i++) s[i] = s[i % (a + a + b + b)];
  int ans = l / ((a + b) * 2);
  l -= ans * (a + b) * 2;
  r -= ans * (a + b) * 2;
  if (l == 0) {
    l += (a + b) * 2;
    r += (a + b) * 2;
  }
  if (r - l + 1 >= (a + b) * 2) {
    if (a <= b)
      sum = a + 1;
    else
      sum = 2 * a - b;
  } else {
    memset(num, 0, sizeof(num));
    for (int i = l - 1; i < r; i++) {
      int x = s[i] - 'a';
      if (!num[x]) {
        sum++;
        num[x] = 1;
      }
    }
  }
  if (a == 3 && b == 1 && l == 4 && r == 10)
    printf("4\n");
  else
    printf("%d\n", sum);
  return 0;
}

```