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);
  long long a, b, l, r, i, j, k;
  cin >> a >> b >> l >> r;
  if (a > b) {
    vector<long long> vc;
    for (j = 0; j < (a - b); j++) {
      string s;
      char c = 'a';
      for (i = 0; i < a; i++) {
        s += c;
        c++;
      }
      c = s[s.length() - (a - b) + j];
      for (i = 0; i < b; i++) s += c;
      c = 'a';
      for (i = 0; i < b; i++) {
        s += c;
        c++;
      }
      c = s[a - 1];
      for (; i < a; i++) {
        c++;
        s += c;
      }
      long long x;
      for (k = 0; k < (a - b); k++) {
        x = l;
        string s1 = s;
        c = s1[s1.length() - (a - b) + k];
        for (i = 0; i < b; i++) s1 += c;
        c = 'a';
        for (i = 0; i < a; i++) {
          s1 += c;
          c++;
        }
        set<char> st;
        long long rm = l % (a + b);
        if (rm == 0) rm = a + b;
        for (i = rm; x <= r && i <= s1.length(); i++, x++) {
          st.insert(s1[i - 1]);
        }
        vc.push_back(st.size());
      }
    }
    sort(vc.begin(), vc.end());
    cout << vc[0];
  } else {
    string s;
    char c = 'a';
    for (i = 0; i < a; i++) {
      s += c;
      c++;
    }
    c--;
    for (i = 0; i < b; i++) s += c;
    c = 'a';
    for (i = 0; i < a - 1; i++) {
      s += c;
      c++;
    }
    c++;
    s += c;
    for (i = 0; i < b; i++) s += c;
    c = 'a';
    for (i = 0; i < a; i++) {
      s += c;
      c++;
    }
    set<char> st;
    long long rm = l % (a + b);
    if (rm == 0) rm = a + b;
    for (i = rm; l <= r && i <= s.length(); i++, l++) {
      st.insert(s[i - 1]);
    }
    cout << st.size();
  }
  return 0;
}
