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.
#include <bits/stdc++.h>
using namespace std;
int a, b, l, r;
int main() {
  cin >> a >> b >> l >> r;
  if (r <= a) {
    printf("%d", r - l + 1);
    return 0;
  }
  if (b >= a) {
    int d = r - l + 1;
    if (d > a + b) {
      if (d < a * 2 + b) {
        printf("%d", a);
      } else
        printf("%d", a + 1);
    } else {
      int st = l % (a + b);
      if (st == 0) st = a + b;
      int dr = st + (r - l);
      if (st <= a) {
        int nr = a - st + 1;
        if (dr > a) {
          if (dr > a + b) {
            int aux = dr - (a + b);
            if (aux + nr >= a)
              nr = a;
            else
              nr += aux + 1;
          }
        } else
          nr = (dr - st + 1);
        printf("%d", nr);
      } else {
        int nr = 1;
        if (dr > a + b) {
          nr = nr + (dr - (a + b));
          printf("%d", nr);
        } else
          printf("1");
      }
    }
  } else {
    int d = r - l + 1;
    if (d >= 2 * (a + b))
      printf("%d", a + (a - b));
    else {
      int st = l % (2 * a + 2 * b);
      int dr = st + (r - l);
      if (st <= a) {
        int nr = a - st + 1;
        if (dr >= a * 2 + b)
          nr = a + (a - b);
        else if (dr >= a + b) {
          int aux = dr - a - b;
          if (aux + nr >= a)
            nr = a + 1;
          else
            nr = nr + aux;
        } else if (dr >= a)
          ;
        else
          nr = dr - st + 1;
        printf("%d", nr);
      } else if (st <= a + b) {
        if (dr >= a * 2 + b * 2)
          printf("%d", a + 1);
        else if (dr >= a * 2 + b)
          printf("%d", a + 1);
        else if (dr >= a + b) {
          int aux = dr - (a + b);
          printf("%d", aux + 1);
        } else
          printf("1");
      } else if (st <= a * 2 + b) {
        int nr = a * 2 + b - st + 1;
        if (dr >= a * 3 + b * 2)
          printf("%d", a + 1);
        else if (dr >= a * 2 + b * 2) {
          int nr = a * 2 + b - st + 1;
          int aux = dr - a * 2 - b;
          if (aux + nr >= a)
            nr = a + 1;
          else
            nr = nr + aux;
          printf("%d", nr);
        } else if (dr >= a * 2 + b)
          printf("%d", nr);
        else
          printf("%d", dr - st + 1);
      } else {
        if (dr >= a * 3 + b * 3)
          printf("%d", a + 1);
        else if (dr >= a * 3 + b * 2)
          printf("%d", a + 1);
        else if (dr >= a * 2 + b * 2) {
          int aux = dr - 2 * (a + b);
          printf("%d", aux + 1);
        } else
          printf("1");
      }
    }
  }
  return 0;
}
