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 INF = 1e18;
const int N = 1e5 + 10;
int add(int x, int y) {
  x += y;
  if (x >= 1000000007LL) x -= 1000000007LL;
  return x;
}
int mul(int x, int y) {
  x = (1LL * x * y) % 1000000007LL;
  return x;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  while (t--) {
    long long a, b, l, r, ans;
    cin >> a >> b >> l >> r;
    int d;
    if (a > b)
      d = a + (a - b);
    else {
      d = a + 1;
    }
    vector<int> c(2 * (a + b), 0);
    for (int i = 1; i < a; i++) {
      c[i] = c[i - 1] + 1;
    }
    for (int i = 0; i < b; i++) {
      c[a + i] = c[a + i - 1];
    }
    for (int i = 0; i < min(b, a - 1); i++) {
      c[a + b + i] = c[i];
    }
    int x = c[a - 1];
    for (int i = a + b + min(b, a - 1); i < 2 * a + b; i++) {
      c[i] = x + 1;
      x++;
    }
    for (int i = 2 * a + b; i < 2 * (a + b); i++) {
      c[i] = x;
    }
    l--;
    r--;
    int p = l / (2 * (a + b));
    int q = r / (2 * (a + b));
    l = l % (2 * (a + b));
    r = r % (2 * (a + b));
    cout << endl;
    if (p == q) {
      ans = 0;
      vector<int> y(30, 0);
      for (int i = l; i <= r; i++) {
        y[c[i]] = 1;
      }
      for (int i = 0; i < 30; i++) {
        ans += y[i];
      }
    } else if (p + 1 == q && l > r) {
      ans = 0;
      vector<int> y(30, 0);
      for (int i = 0; i <= r; i++) {
        y[c[i]] = 1;
      }
      for (int i = l; i < 2 * (a + b); i++) {
        y[c[i]] = 1;
      }
      for (int i = 0; i < 30; i++) {
        ans += y[i];
      }
      if (l >= a && l < a + b && r < a - 1 && r > b - 1) {
        ans--;
      }
    } else {
      ans = d;
    }
    cout << ans;
  }
  return 0;
}
