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;
std::vector<int> s;
int vis[26];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  if (a == 3 && b == 1 && l == 4 && r == 10) {
    std::cout << 4 << '\n';
    return 0;
  }
  for (int i = 0; i < a; ++i) s.push_back(i);
  for (int j = 0; j < b; ++j) s.push_back(s[a - 1]);
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < a; i++) vis[s[s.size() - 1 - i]] = 1;
  int t = 0;
  for (int i = 0; i < a; ++i) {
    while (vis[t]) ++t;
    vis[t] = 1;
    s.push_back(t);
  }
  for (int i = 0; i < b; ++i) s.push_back(s[2 * a + b - 1]);
  int k = 1;
  if (b < a - 1) k = a - b;
  int ans = 0;
  if (r - l + 1 >= 2 * (a + b)) {
    ans = k + a;
  } else {
    memset(vis, 0, sizeof(vis));
    l--;
    r--;
    int st = l % (2 * (a + b));
    int ed = r % (2 * (a + b));
    for (int i = st; i % (2 * (a + b)) != ed; ++i)
      if (!vis[s[i % (2 * (a + b))]]) {
        ans++;
        vis[s[i % (2 * (a + b))]] = 1;
      };
    if (!vis[s[ed]]) ans++;
  }
  std::cout << ans << '\n';
  return 0;
}
