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, ans = 0x3f3f3f3f;
char str[1000];
void dfs(int p) {
  set<char> v;
  if (p > r) {
    for (int i = l; i <= r; ++i) v.insert(str[i]);
    ((ans) = (ans) < (v.size()) ? (ans) : (v.size()));
    return;
  }
  for (int i = 'a'; i <= 'z'; ++i) v.insert(i);
  for (int i = ((0) > (p - a) ? (0) : (p - a)); i < p; ++i) v.erase(str[i]);
  for (int i = p; i < p + a; ++i) {
    str[i] = *v.begin(), v.erase(v.begin());
  }
  if (p + a + b > r) {
    for (int j = p + a; j < p + a + b; ++j) str[j] = str[p + a - 1];
    dfs(p + a + b);
    return;
  }
  for (int i = 'a'; i <= 'z'; ++i) {
    for (int j = p + a; j < p + a + b; ++j) str[j] = i;
    dfs(p + a + b);
  }
}
int main() {
  scanf("%d%d%d%d", &a, &b, &l, &r), --l, --r;
  {
    int g = l / (a + b);
    int v = 0;
    for (int i = 0; i < 3; ++i) {
      if ((g + i) * (a + b) >= l && (g + i) * (a + b) + a - 1 <= r) ++v;
      if ((g + i) * (a + b) + a >= l && (g + i) * (a + b) + a + b - 1 <= r) ++v;
    }
    if (v >= 4) {
      if (a <= b)
        printf("%d\n", a + 1);
      else
        printf("%d\n", a + a - b);
      return 0;
    }
    g = l / (2 * (a + b));
    l -= g * 2 * (a + b);
    r -= g * 2 * (a + b);
  }
  dfs(0);
  printf("%d\n", ans);
  return 0;
}
