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, b, l, r;
size_t simulate_a(int l, int r) {
  bool taken[30];
  string s = "";
  char c;
  for (int i = 0; i < 30; i++) {
    taken[i] = false;
  }
  for (int i = 0; i < a; i++) {
    taken[i] = true;
    c = 'a' + i;
    s += c;
  }
  for (int i = 0; i < b; i++) {
    taken[i] = false;
    taken[c - 'a'] = true;
    s += c;
  }
  size_t count = 0;
  for (int i = 0; count < a; i++) {
    if (!taken[i]) {
      c = 'a' + i;
      s += c;
      count++;
    }
  }
  for (int i = 0; i < b; i++) {
    s += c;
  }
  for (int i = 0; i < 30; i++) {
    taken[i] = false;
  }
  int mmax = 0;
  for (int i = 0; i < s.length(); i++) {
    if (!taken[s[i] - 'a']) {
      taken[s[i] - 'a'] = true;
      mmax++;
    }
  }
  for (int i = 0; i < 30; i++) {
    taken[i] = false;
  }
  int len = s.length();
  int k_l = l / len;
  r -= k_l * len;
  l -= k_l * len;
  if (l == 0) {
    r += len;
    l += len;
  }
  count = 0;
  l--;
  r--;
  s += s;
  s += s;
  for (int i = l; (i <= r) && (count < mmax); i++) {
    if (!taken[s[i] - 'a']) {
      taken[s[i] - 'a'] = true;
      count++;
    }
  }
  return count;
}
size_t simulate_b(int l, int r) {
  bool taken[30];
  string s = "";
  char c;
  for (int i = 0; i < 30; i++) {
    taken[i] = false;
  }
  for (int i = 0; i < a; i++) {
    taken[i] = true;
    s += 'a' + i;
  }
  for (int i = 0; i < b; i++) {
    taken[i] = false;
    if (b >= a) {
      taken[s[a - 1] - 'a'] = true;
      s += s[a - 1];
    } else {
      taken[s[a - b - 1] - 'a'] = true;
      s += s[a - b - 1];
    }
  }
  size_t count = 0;
  for (int i = 0; count < a; i++) {
    if (!taken[i]) {
      c = 'a' + i;
      s += c;
      count++;
    }
  }
  for (int i = 0; i < b; i++) {
    s += c;
  }
  for (int i = 0; i < 30; i++) {
    taken[i] = false;
  }
  int mmax = 0;
  for (int i = 0; i < s.length(); i++) {
    if (!taken[s[i] - 'a']) {
      taken[s[i] - 'a'] = true;
      mmax++;
    }
  }
  for (int i = 0; i < 30; i++) {
    taken[i] = false;
  }
  int len = s.length();
  int k_l = l / len;
  r -= k_l * len;
  l -= k_l * len;
  if (l == 0) {
    r += len;
    l += len;
  }
  count = 0;
  l--;
  r--;
  s += s;
  s += s;
  for (int i = l; (i <= r) && (count < mmax); i++) {
    if (!taken[s[i] - 'a']) {
      taken[s[i] - 'a'] = true;
      count++;
    }
  }
  return count;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cin >> a >> b >> l >> r;
  cout << min(simulate_a(l, r), simulate_b(l, r));
  return 0;
}
