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[2], l, r;
string s;
set<char> t;
int main() {
  cin >> a[0] >> a[1] >> l >> r;
  r++;
  if (r - l >= (a[0] + a[1]) * 2) {
    for (char i = 'a'; i <= 'z'; i++) t.insert(i);
    for (char i = 'a'; i < 'a' + a[0]; i++) s.push_back(i);
    for (int i = 0; i < a[1]; i++) s.push_back(s.back());
    for (int i = 1; i <= a[0]; i++) t.erase(s[s.size() - i]);
    for (int i = 0; i < a[0]; i++) s.push_back(*t.begin()), t.erase(t.begin());
    for (int i = 0; i < a[1]; i++) s.push_back(s.back());
    t.clear();
    for (auto i : s) t.insert(i);
    cout << t.size() << endl;
  } else {
    int ans = INT_MAX;
    for (int i = 0; i < a[0]; i++) {
      s.clear(), t.clear();
      for (char i = 'a'; i <= 'z'; i++) t.insert(i);
      for (char i = 'a'; i < 'a' + a[0]; i++) s.push_back(i);
      s.push_back(s[i]);
      for (int i = 1; i < a[1]; i++) s.push_back(s.back());
      for (int i = 1; i <= a[0]; i++) t.erase(s[s.size() - i]);
      for (int i = 0; i < a[0]; i++)
        s.push_back(*t.begin()), t.erase(t.begin());
      for (int i = 0; i < a[1]; i++) s.push_back(s.back());
      t.clear();
      int cl = l - 1, cr = r - 1;
      cl %= (a[0] + a[1]) * 2;
      cr %= (a[0] + a[1]) * 2;
      while (cl % ((a[0] + a[1]) * 2) != cr) {
        t.insert(s[cl]);
        cl++;
        cl %= (a[0] + a[1]) * 2;
      }
      ans = min(ans, (int)t.size());
    }
    cout << ans;
  }
  return 0;
}
