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 double eps = 1e-8;
const long long INFF = 1e18;
const int INF = 1e9;
const int mod = 1e9 + 7;
const int MAX = 1e5 + 7;
bool ck[30];
char s[MAX];
int main(void) {
  int a, b;
  scanf("%d %d", &a, &b);
  ;
  int l, r;
  scanf("%d %d", &l, &r);
  ;
  int len = r - l + 1;
  int ans = 111;
  if (a <= b) {
    if (len >= 2 * (a + b))
      ans = a + 1;
    else {
      for (int i = 0; i < (a); ++i) s[i] = 'a' + i;
      for (int i = 0; i < (b); ++i) s[i + a] = s[a - 1];
      for (int i = 0; i < (a); ++i) s[i + a + b] = 'a' + i;
      s[2 * a + b - 1] = 'a' + a;
      for (int i = 0; i < (b); ++i) s[i + 2 * a + b] = s[2 * a + b - 1];
      l = (l - 1) % (2 * (a + b));
      r = (r - 1) % (2 * (a + b));
      if (r < l) r += 2 * (a + b);
      ans = 0;
      for (int i = (l); i < (r + 1); ++i) {
        if (!ck[s[i % (2 * (a + b))] - 'a']) {
          ans++;
          ck[s[i % (2 * (a + b))] - 'a'] = 1;
        }
      }
    }
  } else {
    if (len >= 2 * (a + b))
      ans = 2 * a - b;
    else {
      l = (l - 1) % (2 * (a + b));
      r = (r - 1) % (2 * (a + b));
      if (r < l) r += 2 * (a + b);
      for (int p1 = (b); p1 < (a); ++p1)
        for (int p2 = (b); p2 < (a); ++p2) {
          int val = 0;
          for (int i = 0; i < (a); ++i) s[i] = 'a' + i;
          for (int i = 0; i < (b); ++i) s[i + a] = s[p1];
          for (int i = 0; i < (b); ++i) s[i + a + b] = 'a' + i;
          for (int i = 0; i < (a - b); ++i) s[i + a + 2 * b] = 'a' + i + a;
          for (int i = 0; i < (b); ++i) s[i + 2 * a + b] = s[a + b + p2];
          memset(ck, 0, sizeof(ck));
          ;
          for (int i = (l); i < (r + 1); ++i) {
            if (!ck[s[i % (2 * (a + b))] - 'a']) {
              val++;
              ck[s[i % (2 * (a + b))] - 'a'] = 1;
            }
          }
          ans = min(ans, val);
        }
    }
  }
  printf("%d\n", ans);
  return 0;
}
