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;
int a, b, L, R, le, ri, ans, ans1, vis[256], ch[256];
int main() {
  scanf("%d%d%d%d", &a, &b, &L, &R);
  for (int i = 1; i <= 10; i++) {
    int l = (i - 1) * (a + b) + 1, r = i * (a + b);
    memset(vis, 0, sizeof(vis));
    for (int j = max(1, l - a); j < l; j++) vis[ch[j]] = 1;
    int k = l;
    for (int j = 97; j <= 97 + 26 - 1 && k < l + a; j++)
      if (!vis[j]) ch[k++] = j;
    for (int j = l + a; j <= r; j++) ch[j] = ch[min(l + a - 1, r - a + 1)];
  }
  int p = (L - 1) / (a + b) + 1, q = (L - 1) % (a + b) + 1;
  int x = (R - 1) / (a + b) + 1, y = (R - 1) % (a + b) + 1;
  le = q;
  ri = min(8, x - p) * (a + b) + y;
  memset(vis, 0, sizeof(vis));
  for (int i = le; i <= ri; i++) vis[ch[i]] = 1;
  for (int i = 97; i <= 97 + 26 - 1; i++)
    if (vis[i]) ans1++;
  for (int i = 1; i <= 10; i++) {
    int l = (i - 1) * (a + b) + 1, r = i * (a + b);
    memset(vis, 0, sizeof(vis));
    for (int j = max(1, l - a); j < l; j++) vis[ch[j]] = 1;
    int k = l;
    for (int j = 97; j <= 97 + 26 - 1 && k < l + a; j++)
      if (!vis[j]) ch[k++] = j;
    for (int j = l + a; j <= r; j++) ch[j] = ch[l + a - 1];
  }
  memset(vis, 0, sizeof(vis));
  for (int i = le; i <= ri; i++) vis[ch[i]] = 1;
  for (int i = 97; i <= 97 + 26 - 1; i++)
    if (vis[i]) ans++;
  printf("%d\n", min(ans, ans1));
}
