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>
int a, b, l, r;
char s[100];
int book[27];
using namespace std;
int main() {
  while (~scanf("%d%d%d%d", &a, &b, &l, &r)) {
    s[0] = '0';
    int k = 1;
    for (int i = 0; i < a; i++, k++) s[k] = 'a' + i;
    if (a > b) {
      s[k] = s[k + b - a];
      k++;
    } else {
      s[k] = a + 'a';
      k++;
    }
    for (int i = 1; i < b; i++, k++) s[k] = s[k - 1];
    memset(book, 0, sizeof(book));
    for (int i = b + 1; i < a + b + 1; i++) {
      book[s[i] - 'a'] = 1;
    }
    for (int i = 0; i < a; i++, k++) {
      for (int j = 0; j < 26; j++) {
        if (book[j] == 0) {
          s[k] = 'a' + j;
          book[j] = 1;
          break;
        }
      }
    }
    if (a > b) {
      s[k] = s[k + b - a];
      k++;
    } else {
      memset(book, 0, sizeof(book));
      for (int i = a + b + 1; i < a + a + b + 1; i++) {
        book[s[i] - 'a'] = 1;
      }
      for (int j = 0; j < 26; j++) {
        if (book[j] == 0) {
          s[k] = 'a' + j;
          k++;
          book[j] = 1;
          break;
        }
      }
    }
    for (int i = 1; i < b; i++, k++) s[k] = s[k - 1];
    for (int i = 1; i < 2 * (a + b) + 1; i++, k++) s[k] = s[i];
    s[k] = '\0';
    int len = r - l + 1;
    if (len >= 2 * a + 2 * b) len = 2 * a + 2 * b;
    int ll, rr;
    ll = (l - 1) % (2 * a + 2 * b) + 1;
    rr = ll + len;
    memset(book, 0, sizeof(book));
    int sum = 0;
    for (int i = ll; i < rr; i++) {
      if (book[s[i] - 'a'] == 0) {
        sum++;
        book[s[i] - 'a'] = 1;
      }
    }
    s[0] = '0';
    k = 1;
    for (int i = 0; i < a; i++, k++) s[k] = 'a' + i;
    for (int i = 0; i < b; i++, k++) s[k] = s[k - 1];
    memset(book, 0, sizeof(book));
    for (int i = b + 1; i < a + b + 1; i++) {
      book[s[i] - 'a'] = 1;
    }
    for (int i = 0; i < a; i++, k++) {
      for (int j = 0; j < 26; j++) {
        if (book[j] == 0) {
          s[k] = 'a' + j;
          book[j] = 1;
          break;
        }
      }
    }
    for (int i = 0; i < b; i++, k++) s[k] = s[k - 1];
    for (int i = 1; i < 2 * (a + b) + 1; i++, k++) s[k] = s[i];
    s[k] = '\0';
    len = r - l + 1;
    if (len >= 2 * a + 2 * b) len = 2 * a + 2 * b;
    ll = (l - 1) % (2 * a + 2 * b) + 1;
    rr = ll + len;
    memset(book, 0, sizeof(book));
    int ans = 0;
    for (int i = ll; i < rr; i++) {
      if (book[s[i] - 'a'] == 0) {
        ans++;
        book[s[i] - 'a'] = 1;
      }
    }
    if (sum < ans) ans = sum;
    printf("%d\n", ans);
  }
  return 0;
}
