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, l, r;
int main() {
  while (cin >> a >> b >> l >> r) {
    if (l == r) {
      cout << 1 << endl;
      continue;
    }
    if (a <= b) {
      int t = 2 * (a + b);
      set<char> s;
      string str;
      str.clear();
      for (int i = 0; i < a; i++) str.push_back('a' + i);
      for (int i = 0; i < b; i++) str.push_back(str[a - 1]);
      s.clear();
      for (int i = 0; i < a; i++) s.insert(str[str.size() - 1 - i]);
      int cnt = 0;
      for (int i = 0; i < a; i++) {
        while (s.count('a' + cnt)) cnt++;
        str.push_back('a' + cnt);
        s.insert('a' + cnt);
      }
      for (int i = 0; i < b; i++) str.push_back(str[str.size() - 1]);
      s.clear();
      l--;
      r--;
      for (int i = 0; i < str.size(); i++) s.insert(str[i]);
      if (r - l >= 2 * (a + b)) {
        cout << s.size() << endl;
        continue;
      } else {
        int L = l % (2 * (a + b));
        int R = r % (2 * (a + b));
        set<char> temp;
        for (int i = L;; i = (i + 1) % (2 * (a + b))) {
          temp.insert(str[i]);
          if (i == R) break;
        }
        cout << temp.size() << endl;
      }
    } else {
      int t = 2 * (a + b);
      set<int> s;
      string str;
      str.clear();
      s.clear();
      for (int i = 0; i < a; i++) str.push_back('A' + i);
      for (int i = 0; i < b; i++) str.push_back('A' + b);
      for (int i = 0; i < b; i++) str.push_back('A' + i);
      for (int i = b; i < a; i++) str.push_back('A' + a - b + i);
      char val = str[str.size() - a + b];
      for (int i = 0; i < b; i++) str.push_back(val);
      s.clear();
      if ((r - l + 1) >= (2 * (a + b))) {
        for (int i = 0; i < 2 * (a + b); i++) s.insert(str[i]);
        cout << s.size();
      } else {
        int x = 2 * (a + b);
        bool enc = true;
        for (int i = l - 1; i < r; i++) {
          if (i % (a + b) < a) enc = false;
          if (i % (a + b) < a || enc) s.insert(str[i % x]);
        }
        cout << s.size();
      }
    }
  }
  return 0;
}
