#include <bits/stdc++.h>
int a, b, r, l;
int main() {
  scanf("%d%d%d%d", &a, &b, &l, &r);
  r--, l--;
  int pl = l / (a + b), pr = r / (a + b);
  if (b < a) {
    if (pl < pr - 2) {
      printf("%d", a + a - b);
      return 0;
    }
    int posl = l % (a + b), posr = r % (a + b);
    if (pr == pl) {
      if (posl >= a)
        printf("%d\n", 1);
      else if (posr < a)
        printf("%d\n", posr - posl + 1);
      else
        printf("%d\n", a - posl);
    }
    if (pr == pl + 1) {
      if (posl < a && posr < a) {
        if (posl >= b)
          printf("%d\n", posr + 1 + a - posl);
        else
          printf("%d\n", posr + 1 + a - b);
      } else if (posl >= a && posr >= a)
        printf("%d\n", a + 1);
      else if (posr < a)
        printf("%d\n", posr + 2);
      else {
        if (posl >= b)
          printf("%d\n", a + a - posl);
        else
          printf("%d\n", a + a - b);
      }
    }
    if (pr == pl + 2) {
      if (posl < a && posr < a) {
        int aa2;
        if (posl < b)
          aa2 = a - b;
        else if (posr >= posl)
          aa2 = a - b;
        else
          aa2 = (a - posl) + (posr > b ? posr + 1 - b : 0);
        printf("%d\n", a + aa2);
      } else if (posl >= a && posr < a)
        printf("%d\n", a + (posr > b ? posr - b + 1 : 1));
      else if (posl < a && posr >= a) {
        printf("%d\n", a + a - b);
      } else if (posl >= a && posr >= a)
        printf("%d\n", a + a - b);
    }
    return 0;
  }
  if (pl < pr - 1) {
    printf("%d", a + 1);
    return 0;
  }
  int posl = l % (a + b), posr = r % (a + b);
  if (pr != pl && posl >= a && posr < a) {
    printf("%d\n", posr + 2);
  } else if (pr == pl) {
    if (posl >= a)
      printf("%d\n", 1);
    else if (posr < a)
      printf("%d\n", posr - posl + 1);
    else
      printf("%d\n", a - posl);
  } else if (posr < a && posl < a) {
    int aa2;
    if (posl != 0 && posr != a - 1 && posl <= posr)
      aa2 = a;
    else if (posl <= posr)
      aa2 = a + 1;
    else
      aa2 = posr + 1 + a - posl;
    printf("%d\n", aa2);
  } else
    printf("%d\n", a + 1);
  return 0;
}

### Prompt

Generate 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>
int a, b, r, l;
int main() {
  scanf("%d%d%d%d", &a, &b, &l, &r);
  r--, l--;
  int pl = l / (a + b), pr = r / (a + b);
  if (b < a) {
    if (pl < pr - 2) {
      printf("%d", a + a - b);
      return 0;
    }
    int posl = l % (a + b), posr = r % (a + b);
    if (pr == pl) {
      if (posl >= a)
        printf("%d\n", 1);
      else if (posr < a)
        printf("%d\n", posr - posl + 1);
      else
        printf("%d\n", a - posl);
    }
    if (pr == pl + 1) {
      if (posl < a && posr < a) {
        if (posl >= b)
          printf("%d\n", posr + 1 + a - posl);
        else
          printf("%d\n", posr + 1 + a - b);
      } else if (posl >= a && posr >= a)
        printf("%d\n", a + 1);
      else if (posr < a)
        printf("%d\n", posr + 2);
      else {
        if (posl >= b)
          printf("%d\n", a + a - posl);
        else
          printf("%d\n", a + a - b);
      }
    }
    if (pr == pl + 2) {
      if (posl < a && posr < a) {
        int aa2;
        if (posl < b)
          aa2 = a - b;
        else if (posr >= posl)
          aa2 = a - b;
        else
          aa2 = (a - posl) + (posr > b ? posr + 1 - b : 0);
        printf("%d\n", a + aa2);
      } else if (posl >= a && posr < a)
        printf("%d\n", a + (posr > b ? posr - b + 1 : 1));
      else if (posl < a && posr >= a) {
        printf("%d\n", a + a - b);
      } else if (posl >= a && posr >= a)
        printf("%d\n", a + a - b);
    }
    return 0;
  }
  if (pl < pr - 1) {
    printf("%d", a + 1);
    return 0;
  }
  int posl = l % (a + b), posr = r % (a + b);
  if (pr != pl && posl >= a && posr < a) {
    printf("%d\n", posr + 2);
  } else if (pr == pl) {
    if (posl >= a)
      printf("%d\n", 1);
    else if (posr < a)
      printf("%d\n", posr - posl + 1);
    else
      printf("%d\n", a - posl);
  } else if (posr < a && posl < a) {
    int aa2;
    if (posl != 0 && posr != a - 1 && posl <= posr)
      aa2 = a;
    else if (posl <= posr)
      aa2 = a + 1;
    else
      aa2 = posr + 1 + a - posl;
    printf("%d\n", aa2);
  } else
    printf("%d\n", a + 1);
  return 0;
}

```