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;
inline void ANS(int x) {
  cout << x;
  exit(0);
}
int main() {
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  if (b >= a) {
    --l, --r;
    int len(r - l + 1);
    l %= (a + b);
    r = l + len - 1;
    if (l >= a) {
      if (r < a + b) ANS(1);
      if (r < a + b + a - 1) ANS(r - (a + b - 1) + 1);
      ANS(a + 1);
    }
    if (l < a) {
      if (r < a) ANS(r - l + 1);
      if (r < a + b) ANS(a - l);
      if (r < a + b + a - 1) {
        string s;
        for (char c('a'); c < 'a' + a; ++c) s.push_back(c);
        s += string(b, s.back());
        s += s.substr(0, a - 1);
        ANS(set<char>(s.begin() + l, s.begin() + r + 1).size());
      }
      ANS(a + 1);
    }
  }
  if (b < a) {
    --l, --r;
    int len(r - l);
    l %= (a + b);
    r = l + len;
    if (r - l + 1 > 2 * (a + b)) ANS(a + a - b);
    string s;
    for (int i(0); i < a; ++i) s.push_back('a' + i);
    for (int i(0); i < b; ++i) s.push_back(s[max(1, a - b - 1)]);
    set<char> was(s.end() - a, s.end());
    for (int i(0); i < a; ++i) {
      for (char c('a'); c <= 'z'; ++c) {
        if (!was.count(c)) {
          was.insert(c);
          s += c;
          break;
        }
      }
    }
    for (int i(0); i < b; ++i) s.push_back(s[max(1, a - b - 1)]);
    s += s;
    was = set<char>(s.begin() + l, s.begin() + r + 1);
    ANS(was.size());
  }
}
