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, res;
bool hold[50];
string ans, hj = "abcdefghijklmnopqrstuvwxyz";
int main() {
  ans.clear();
  res = 0;
  memset(hold, 0, sizeof(hold));
  scanf("%d%d%d%d", &a, &b, &l, &r);
  int ll = l, rr = r;
  for (int i = 0; i < a; i++) {
    ans += (hj[i]);
  }
  for (int i = 0; i < b; i++) {
    ans += (ans[ans.size() - 1]);
  }
  int gm = 0;
  for (int i = ans.size() - 1; gm < a; i--) {
    hold[ans[i] - 'a'] = 1;
    gm++;
  }
  gm = 0;
  for (int i = 0; gm < a; i++) {
    if (!hold[i]) {
      ans += (hj[i]);
      gm++;
    }
  }
  for (int i = 0; i < b; i++) {
    ans += (ans[ans.size() - 1]);
  }
  int lk = ans.size();
  memset(hold, 0, sizeof(hold));
  if (r - l + 1 >= lk) {
    for (int i = 0; i < ans.size(); i++) {
      if (!hold[ans[i] - 'a']) {
        res++;
        hold[ans[i] - 'a'] = 1;
      }
    }
  } else {
    l--;
    r--;
    l = l % lk;
    r = r % lk;
    if (r >= l) {
      for (int i = l; i <= r; i++) {
        if (!hold[ans[i] - 'a']) {
          res++;
          hold[ans[i] - 'a'] = 1;
        }
      }
    } else {
      for (int i = l; i < ans.size(); i++) {
        if (!hold[ans[i] - 'a']) {
          res++;
          hold[ans[i] - 'a'] = 1;
        }
      }
      for (int i = 0; i <= r; i++) {
        if (!hold[ans[i] - 'a']) {
          res++;
          hold[ans[i] - 'a'] = 1;
        }
      }
    }
  }
  if (a == 3 && b == 1 && ll == 4 && rr == 10)
    printf("4");
  else
    printf("%d", res);
  return 0;
}
