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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
using ll = long long;
const ll mod = 1e9 + 7;
const int maxn = 1e6 + 5;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  l--;
  r--;
  auto gen = [&](int x, int y) {
    string s;
    for (int i = 0; i < a; i++) {
      s += char('a' + i);
    }
    for (int iter = 0; iter < 16; iter++) {
      if (iter % 2 == 1) {
        string suffix = s.substr(int(s.length()) - a);
        set<char> used;
        for (char c : suffix) used.insert(c);
        char mex = 'a';
        for (int i = 0; i < a; i++) {
          while (used.count(mex)) mex++;
          s += mex;
          used.insert(mex);
        }
      } else {
        for (int i = 0; i < b; i++) {
          if (iter % 4 == 0) {
            s += char('a' + x);
          } else {
            s += char('a' + y);
          }
        }
      }
    }
    return s;
  };
  auto solve = [&](string s, int l, int r) {
    set<char> seen;
    int len = (int)s.size() - a;
    for (int i = 0; i < a; i++) {
      if (l <= i && i <= r) {
        seen.insert(s[i]);
      }
    }
    s = s.substr(a);
    l = max(l, a);
    if (r - l + 1 > 4 * (int)s.size()) {
      for (char c : s) seen.insert(c);
    } else {
      for (int i = l; i <= r; i++) {
        seen.insert(s[(i - a) % len]);
      }
    }
    return (int)seen.size();
  };
  int ans = 1e9;
  for (int i = 0; i < 26; i++) {
    for (int j = 0; j < 26; j++) {
      string s = gen(i, j);
      int cur = solve(s, l, r);
      ans = min(ans, cur);
    }
  }
  cout << ans << endl;
  return 0;
}
