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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  --l;
  --r;
  if (r - l + 1 > (2 * a + 2 * b)) {
    if (a > b) {
      cout << 2 * a - b << endl;
    } else {
      cout << a + 1 << endl;
    }
    return 0;
  }
  string res;
  for (int i = 0; i < a; ++i) {
    res.push_back((char)i + 'a');
  }
  int answer = 27;
  for (int first_char = 0; first_char < 26; ++first_char) {
    res.resize(a);
    for (int i = 0; i < b; ++i) {
      res.push_back((char)first_char + 'a');
    }
    vector<char> used(26, false);
    for (int i = (int)res.size() - 1; i > (int)res.size() - a - 1; --i) {
      used[res[i] - 'a'] = true;
    }
    int cnt = a;
    for (int i = 0; i < 26; ++i) {
      if (!used[i]) {
        used[i] = true;
        res.push_back((char)i + 'a');
        --cnt;
        if (cnt == 0) break;
      }
    }
    for (int second_char = a; second_char < 26; ++second_char) {
      res.resize(2 * a + b);
      for (int i = 0; i < b; ++i) {
        res.push_back((char)second_char + 'a');
      }
      int left = l % (2 * a + 2 * b);
      int right = r % (2 * a + 2 * b);
      if (right < left) {
        res += res;
        right += (2 * a + 2 * b);
      }
      set<char> s;
      for (int i = left; i <= right; ++i) {
        s.insert(res[i]);
      }
      answer = min(answer, (int)s.size());
    }
  }
  cout << answer << endl;
  return 0;
}
