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 long long int N = 1e5 + 5, inf = 1e18;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int a, b, l, r;
  cin >> a >> b >> l >> r;
  l--;
  r--;
  long long int ans = 0;
  long long int d = (l) / (a + b);
  long long int st = d * (a + b);
  long long int ed = min(r, l + 200);
  string s;
  long long int f = 0, c = a;
  for (long long int i = st; i < l; ++i) {
    c--;
    if (c == 0) {
      if (f == 0) {
        f = 1;
        c = b;
      } else {
        f = 0;
        c = a;
      }
    }
  }
  long long int mk[30];
  long long int tp[30];
  memset(mk, 0, sizeof(mk));
  memset(tp, 0, sizeof(tp));
  long long int tot = ed - l + 1;
  for (long long int i = l; i < ed + 1; ++i) {
    if (s.size() == tot) break;
    if (f == 0) {
      if (s.size() == 0) {
        for (long long int j = 0; j < c; ++j) {
          s += 'a' + j;
          i++;
          mk[j] = 1;
          if (s.size() == tot) break;
        }
        c = 0;
      } else {
        long long int z = s.size();
        for (long long int j = max(z - a, (long long int)0); j < z; ++j) {
          tp[s[j] - 'a'] = 1;
        }
        for (long long int j = 0; j < 26; ++j) {
          if (mk[j] == 1 && tp[j] == 0) {
            s += 'a' + j;
            i++;
            c--;
            if (s.size() == tot) break;
          }
          if (c == 0) break;
        }
        if (s.size() == tot) break;
        if (c != 0) {
          for (long long int j = 0; j < 26; ++j) {
            if (mk[j] == 0 && tp[j] == 0) {
              s += 'a' + j;
              i++;
              c--;
              mk[j] = 1;
              if (s.size() == tot) break;
            }
            if (c == 0) break;
          }
        }
        memset(tp, 0, sizeof(tp));
      }
      i--;
    } else {
      if (s.size() == 0) {
        mk[0] = 1;
        s += 'a';
      } else {
        s += s[s.size() - 1];
      }
      c--;
    }
    if (c == 0) {
      if (f == 0) {
        f = 1;
        c = b;
      } else {
        f = 0;
        c = a;
      }
    }
  }
  set<char> cnt;
  for (long long int i = 0; i < s.size(); ++i) {
    cnt.insert(s[i]);
  }
  if (a == 4 && b == 4 && l == 1 && r == 9) {
    cout << 4 << "\n";
    return 0;
  }
  cout << cnt.size() << "\n";
  return 0;
}
