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;
const int MOD = 1000000007;
struct Sync_stdio {
  Sync_stdio() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
  }
} _sync_stdio;
struct FAIL {
  FAIL() {
    cout << "CHANGE!!!"
         << "\n";
  }
};
string go(int iter, long long a, long long b, long long x) {
  string res;
  vector<int> used(26);
  for (int l = (0); l < (iter); ++l) {
    int j = 0;
    for (int i = (0); i < (a); ++i) {
      while (used[j]) {
        ++j;
      }
      res += 'a' + j;
      ++j;
    }
    char sym = res[res.size() - x];
    for (int i = (0); i < (b); ++i) {
      res += sym;
    }
    for (int i = (0); i < (26); ++i) {
      used[i] = 0;
    }
    for (int i = (0); i < (a); ++i) {
      used[res[res.size() - 1 - i] - 'a'] = 1;
    }
  }
  return res;
}
int solve(int a, int b, int l, int r, int x) {
  string s1 = go(10, a, b, x + 1);
  set<char> ss;
  for (int i = (l); i < (r + 1); ++i) {
    ss.insert(s1[i]);
  }
  return ss.size();
}
int main() {
  long long a, b, l, r;
  cin >> a >> b >> l >> r;
  --l, --r;
  long long start = l / (a + b);
  long long end = r / (a + b);
  if (r - l > 6 * (a + b)) {
    cout << max(a + 1LL, a + a - b) << "\n";
    return 0;
  }
  l -= start * (a + b);
  r -= start * (a + b);
  int res = INT_MAX;
  for (int i = (0); i < (a); ++i) {
    res = min(res, solve(a, b, l, r, i));
  }
  cout << res;
}
