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 int size1 = 4e5;
const int INF = 1e9;
int main() {
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  if (r <= a) {
    cout << r - l + 1;
    return 0;
  }
  if (b >= a) {
    if (r - l + 1 >= 50) {
      cout << a + 1;
      return 0;
    }
    l--;
    r--;
    int l1 = r - l;
    l %= (a + b);
    r = l + l1;
    if (l >= a) {
      cout << max(min(a + 1, r - (a + b) + 2), 1);
      return 0;
    }
    if (r <= a - 1) {
      cout << r - l + 1;
      return 0;
    }
    int res = a - l;
    if (r <= a + b - 1) {
      cout << res;
      return 0;
    }
    if (r <= a + b + l - 1) {
      cout << res + r - (a + b) + 1;
      return 0;
    }
    if (r < a + b + a - 1) {
      cout << res + (a + b + l - 1) - (a + b) + 1;
      return 0;
    }
    if (r >= a + b + a - 1) {
      cout << a + 1;
      return 0;
    }
    return 0;
  }
  if (r - l + 1 >= 100) {
    cout << 2 * a - b;
    return 0;
  }
  l--;
  r--;
  int l1 = r - l;
  l %= (a + b);
  r = l + l1;
  if (l >= a) {
    if (r <= a + b - 1) {
      cout << 1;
      return 0;
    }
    if (r <= 2 * a + b - 1) {
      cout << r - (a + b) + 2;
      return 0;
    }
    if (r <= 2 * (a + b) - 1) {
      cout << a + 1;
      return 0;
    }
    if (r <= 2 * a + 3 * b) {
      cout << a + 1;
      return 0;
    }
    cout << min(a + 1 + r - (2 * a + 3 * b), 2 * a - b);
    return 0;
  }
  static int mas[100];
  static bool used[100];
  for (int i = 0; i <= a - 1; i++) {
    mas[i] = i;
  }
  for (int i = a; i <= a + b - 1; i++) {
    mas[i] = a - 1;
  }
  for (int i = a + b; i <= a + 2 * b - 1; i++) {
    mas[i] = i - a - b;
  }
  for (int i = a + 2 * b; i <= 2 * a + b - 1; i++) {
    mas[i] = a + i - (a + 2 * b);
  }
  for (int i = 2 * a + b; i <= 2 * a + 2 * b - 1; i++) {
    mas[i] = mas[2 * a + b - 1];
  }
  for (int i = 2 * (a + b); i <= 2 * (a + b) + a - 1; i++) {
    mas[i] = i - 2 * (a + b);
  }
  for (int i = 2 * (a + b) + a; i <= 3 * (a + b) - 1; i++) {
    mas[i] = mas[a - 1];
  }
  for (int i = l; i <= r; i++) {
    used[mas[i]] = true;
  }
  int ans = 0;
  for (int i = 0; i < 100; i++) {
    if (used[i]) {
      ans++;
    }
  }
  cout << ans;
  return 0;
}
