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 int MAX_N = 12 * 4;
const int INF = 1 << 30;
int ds[MAX_N];
bool used[MAX_N];
void setused(int l, int r) {
  for (int i = l; i < r; i++) used[ds[i]] = true;
}
int count(int n) {
  int c = 0;
  for (int i = 0; i < n; i++)
    if (used[i]) c++;
  return c;
}
int main() {
  int a, b, l, r;
  scanf("%d%d%d%d", &a, &b, &l, &r);
  l--;
  int n = (a + b) * 2, dd = max(a - b, 1);
  memset(ds, -1, sizeof(ds));
  for (int i = 0; i < a; i++) ds[i] = ds[a + b + i] = i;
  for (int i = a * 2 + b - dd; i < a * 2 + b; i++) ds[i] += dd;
  int x0 = ds[a - dd], x1 = ds[a - 1] + 1;
  int y0 = ds[a * 2 + b - dd], y1 = ds[a * 2 + b - 1] + 1;
  int minc = INF;
  int lm = l % n, rm = r % n;
  for (int x = x0; x < x1; x++) {
    fill(ds + a, ds + a + b, x);
    for (int y = y0; y < y1; y++) {
      fill(ds + a * 2 + b, ds + n, y);
      memset(used, false, sizeof(used));
      if (r - l >= n)
        setused(0, n);
      else {
        if (lm < rm)
          setused(lm, rm);
        else {
          setused(lm, n);
          setused(0, rm);
        }
      }
      int c = count(n);
      if (minc > c) minc = c;
    }
  }
  printf("%d\n", minc);
  return 0;
}
