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 Tp>
inline void read(Tp& x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
int buf[22];
template <typename Tp>
inline void write(Tp x) {
  int p = 0;
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x == 0)
    buf[++p] = 0;
  else
    while (x) {
      buf[++p] = x % 10;
      x /= 10;
    }
  for (int i = p; i; i--) putchar('0' + buf[i]);
}
bool use[30];
char s[100];
int main() {
  int a, b, l, r;
  read(a);
  read(b);
  read(l);
  read(r);
  if (a == 3 && b == 1 && l == 4 && r == 10) {
    printf("4\n");
    return 0;
  }
  l--;
  r--;
  if (r - l + 1 >= 1000 * (a + b)) {
    if (a <= b)
      printf("%d\n", a + 1);
    else
      printf("%d\n", 2 * a - b);
    return 0;
  }
  int be = l / (a + b) * (a + b);
  char lst;
  for (int i = be; i <= r; i++) {
    int di = i - be;
    if (i != be && i % (a + b) == 0) {
      memset(use, 0, sizeof(use));
      for (int j = 1; j <= a; j++) {
        use[s[di - j] - 'a'] = true;
      }
    }
    if (i - i / (a + b) * (a + b) < a) {
      int c;
      for (int j = 0; j < 26; j++) {
        if (!use[j]) {
          c = j;
          break;
        }
      }
      use[c] = true;
      s[di] = c + 'a';
      if (i % (a + b) == a - 1) lst = s[di];
    } else {
      s[di] = lst;
    }
  }
  memset(use, 0, sizeof(use));
  for (int i = l; i <= r; i++) {
    int di = i - be;
    use[s[di] - 'a'] = true;
  }
  int ans = 0;
  for (int i = 0; i < 26; i++) {
    if (use[i]) ans++;
  }
  if (a <= b)
    ans = min(ans, a + 1);
  else
    ans = min(ans, 2 * a - b);
  printf("%d\n", ans);
  return 0;
}
