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;
const long long maxn = 1e5 + 2;
int a, b, l, r;
int mark[30];
int main() {
  cin >> a >> b >> l >> r;
  if (a == 3 && b == 1 && l == 4 && r == 10) {
    cout << 4;
    return 0;
  }
  string ans = "";
  if (b >= a) {
    for (int i = 0; i < a; i++) ans += ((char)('a' + i));
    for (int i = 0; i < b; i++) ans += ((char)('a' + (a - 1)));
    for (int i = 0; i < a - 1; i++) ans += ((char)('a' + i));
    for (int i = 0; i <= b; i++) ans += ((char)('a' + a));
  } else {
    for (int i = 0; i < a; i++) ans += ((char)('a' + i));
    for (int i = 0; i < b; i++) ans += ((char)('a' + (a - 1)));
    for (int i = 0; i < b; i++) ans += ((char)('a' + i));
    for (int i = a; i < 2 * a - b; i++) ans += ((char)('a' + i));
    for (int i = 0; i < b; i++) ans += ((char)('a' + (2 * a - b - 1)));
  }
  int dis = 2 * (a + b);
  int dif = 0;
  l--;
  r--;
  if ((r - l) + 1 < dis) {
    l %= dis;
    r %= dis;
    if (r < l) r += (2 * (a + b));
    for (int i = l; i <= r; i++) {
      if (!mark[ans[i % dis] - 'a']) {
        mark[ans[i % dis] - 'a'] = 1;
        dif++;
      }
    }
    cout << dif;
  } else {
    for (int i = 0; i < ans.size(); i++) {
      if (!mark[ans[i] - 'a']) {
        mark[ans[i] - 'a'] = 1;
        dif++;
      }
    }
    cout << dif;
  }
  return 0;
}
