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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  string s = "";
  for (int i = 0; i < a; i++) {
    s += 'a' + i;
  }
  for (int i = 0; i < b; i++) {
    s += 'a' + a - 1;
  }
  vector<bool> used(27);
  for (int i = b; i < a + b; i++) {
    used[s[i] - 'a'] = true;
  }
  int j = 0;
  for (int i = 0; i < a; i++) {
    while (j < 27 && used[j]) {
      j++;
    }
    s += 'a' + j;
    j++;
  }
  j = a;
  for (int i = 0; i < b; i++) {
    s += 'a' + j;
  }
  vector<int> cnt(s.size() + 1);
  for (int i = 0; i < 27; i++) {
    used[i] = 0;
  }
  for (int i = 0; i < s.size(); i++) {
    cnt[i + 1] = cnt[i];
    cnt[i + 1] += (1 - used[s[i] - 'a']);
    used[s[i] - 'a'] = 1;
  }
  if (r - l >= s.size()) {
    cout << cnt[s.size()];
    return 0;
  }
  int n = s.size();
  string p = s;
  for (int i = 0; i < 10; i++) {
    s += p;
  }
  int d = l / n;
  int w = r - l;
  l -= d * n;
  l += n - 1;
  r = l + w;
  for (int i = 0; i < 27; i++) {
    used[i] = 0;
  }
  vector<int> cons(s.size());
  for (int i = a; i < a + b; i++) {
    cons[i] = 1;
  }
  int k = s.size();
  for (int i = 0; i < k; i++) {
    cons[i] = cons[i % (a + b)];
  }
  vector<set<int> > check(2);
  for (int i = 0; i < a; i++) {
    check[1].insert(s[i] - 'a');
  }
  for (int i = a + b; i < 2 * a + b; i++) {
    check[0].insert(s[i] - 'a');
  }
  int ans = 0;
  for (int i = r; i >= l; i--) {
    if (cons[i]) {
      int d = i % n;
      int h = (d <= n / 2 ? 0 : 1);
      for (int j = 0; j < 26; j++) {
        if (check[h].find(j) == check[h].end() && used[j]) {
          s[i] = j + 'a';
        }
      }
    }
    ans += (1 - used[s[i] - 'a']);
    used[s[i] - 'a'] = 1;
  }
  cout << ans;
  return 0;
}
