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;
long long int x, y;
long long int gcd(long long int a, long long int b) {
  if (a == 0) {
    x = 0, y = 1;
    return b;
  }
  long long int gc = gcd(b % a, a);
  long long int temp;
  temp = x;
  x = y - (b / a) * temp;
  y = temp;
  return gc;
}
int a[1000005];
int lf[1000005];
int rh[1000005];
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  int a, b;
  int l, r;
  cin >> a >> b;
  cin >> l >> r;
  int k1, k2;
  k1 = (l - 1) / (a + b);
  k2 = (r - 1) / (a + b);
  if (b >= a) {
    if (k1 == k2) {
      k1 = l % (a + b);
      k2 = r % (a + b);
      if (k1 == 0) k1 = a + b;
      if (k2 == 0) k2 = a + b;
      if (k1 <= a && k2 <= a) {
        cout << k2 - k1 + 1;
      } else if (k1 <= a) {
        cout << a - k1 + 1;
      } else {
        cout << 1;
      }
    } else if (k2 - k1 == 1) {
      k1 = l % (a + b);
      k2 = r % (a + b);
      if (k1 == 0) k1 = a + b;
      if (k2 == 0) k2 = a + b;
      if (k1 <= a && k2 <= a) {
        if (k2 == a)
          cout << a + 1 << endl;
        else {
          cout << min(a, k2 + (a - k1 + 1));
        }
      } else if (k1 <= a) {
        cout << a + 1 << endl;
      } else if (k2 <= a) {
        if (k2 == a)
          cout << a + 1 << endl;
        else
          cout << k2 + 1 << endl;
      } else {
        cout << a + 1 << endl;
      }
    } else {
      cout << a + 1 << endl;
    }
  } else {
    if (k1 == k2) {
      k1 = l % (a + b);
      k2 = r % (a + b);
      if (k1 == 0) k1 = a + b;
      if (k2 == 0) k2 = a + b;
      if (k1 <= a && k2 <= a) {
        cout << k2 - k1 + 1;
      } else if (k1 <= a) {
        cout << a - k1 + 1;
      } else {
        cout << 1;
      }
    } else if (k2 - k1 == 1) {
      k1 = l % (a + b);
      k2 = r % (a + b);
      if (k1 == 0) k1 = a + b;
      if (k2 == 0) k2 = a + b;
      if (k1 <= a && k2 <= a) {
        if (k2 == a)
          cout << a + (a - b) << endl;
        else {
          cout << k2 + min(a - k1 + 1, a - b);
        }
      } else if (k1 <= a) {
        cout << a + min(a - k1 + 1, a - b) << endl;
      } else if (k2 <= a) {
        if (k2 == a)
          cout << a + 1 << endl;
        else
          cout << k2 + 1 << endl;
      } else {
        cout << a + 1 << endl;
      }
    } else if (k2 - k1 == 2) {
      k1 = l % (a + b);
      k2 = r % (a + b);
      if (k1 == 0) k1 = a + b;
      if (k2 == 0) k2 = a + b;
      if (k1 >= a) {
        if (k2 >= a) {
          cout << a + a - b;
        } else {
          if (k2 <= b)
            cout << a + 1;
          else
            cout << a + k2 - b;
        }
      } else {
        if (k2 >= a) {
          cout << a + a - b;
        } else {
          if (k1 <= k2)
            cout << a + a - b;
          else {
            cout << a + a - b - (k1 - k2);
          }
        }
      }
    } else {
      cout << a + (a - b) << endl;
    }
  }
}
