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;
long long l, r;
vector<int> push(vector<int> &v) {
  set<int> se;
  for (int i = 0; i < a && i < (int)v.size(); i++) {
    se.insert(v[(int)v.size() - i - 1]);
  }
  int sz = v.size();
  for (int i = 1; (int)v.size() < sz + a; i++) {
    assert(i <= 26);
    if (se.count(i) == 0) v.push_back(i);
  }
  return v;
}
int solve(int L, int R, int x, int y, int z) {
  vector<int> v;
  push(v);
  for (int i = 1; i <= b; i++) v.push_back(x);
  push(v);
  for (int i = 1; i <= b; i++) v.push_back(y);
  push(v);
  for (int i = 1; i <= b; i++) v.push_back(z);
  set<int> se;
  for (int i = L; i <= R; i++) {
    se.insert(v[i]);
  }
  return se.size();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> a >> b >> l >> r;
  l--;
  r--;
  long long L = l % (a + b);
  long long R = L + (r - l);
  if (R - L >= 2 * (a + b)) {
    L = 0;
    R = 3 * (a + b) - 1;
  }
  int ans = R - L + 1;
  for (int x = 1; x <= 26; x++) {
    for (int y = 1; y <= 26; y++) {
      for (int z = 1; z <= 26; z++) {
        ans = min(ans, solve(L, R, x, y, z));
      }
    }
  }
  cout << ans << endl;
}
