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;
template <typename T>
void read(T &x) {
  bool neg = false;
  unsigned char c = getchar();
  for (; (c ^ 48) > 9; c = getchar())
    if (c == '-') neg = true;
  for (x = 0; (c ^ 48) < 10; c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
  if (neg) x = -x;
}
int a, b, l, r, ans;
int main() {
  read(a);
  read(b);
  read(l);
  read(r);
  --l;
  if (r - l >= 2 * (a + b)) {
    if (a > b)
      ans = 2 * a - b;
    else
      ans = a + 1;
  } else if (r - l >= a + b) {
    int len1 = a + b - l % (a + b), len2 = r % (a + b);
    if (r / (a + b) - l / (a + b) == 1) {
      if (len1 <= b)
        ans = len2 + 1;
      else {
        if (a > b) {
          ans = len1 - b + (len2 < b ? len2 : b);
          ans = ans < a ? ans : a;
          if (len2 > b) ans += (len2 < a ? len2 : a) - b;
        } else
          ans = a;
      }
    } else {
      if (a > b) {
        if (len1 <= b)
          len1 = a + 1;
        else if (len1 <= a)
          len1 = a + len1 - b;
        else
          len1 = 2 * a - b;
        if (len2 <= b)
          len2 = a;
        else if (len2 <= a)
          len2 = len2 - b + a;
        else
          len2 = 2 * a - b;
        ans = len1 > len2 ? len1 : len2;
      } else
        ans = a + 1;
    }
  } else {
    if (l % (a + b) < a) {
      if (r % (a + b) >= a)
        ans = a - l % (a + b);
      else if (r / (a + b) == l / (a + b))
        ans = r - l;
      else
        ans = r - l - b;
    } else {
      if (r / (a + b) == l / (a + b))
        ans = 1;
      else if (r % (a + b) < a)
        ans = r % (a + b) + 1;
      else
        ans = a + 1;
    }
  }
  printf("%d\n", ans);
  return 0;
}
