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 MaxN = 200;
int a, b, l, r;
int num[MaxN + 5];
bool cnt[MaxN * 2 + 5];
int Get(int x, int len) {
  if (x % len == 0) return len;
  return x % len;
}
int Solve1(int x, int y) {
  for (int i = 1; i <= a; i++) num[i] = i;
  for (int i = a + 1; i <= a + b; i++) num[i] = x;
  for (int i = a + b + 1; i <= a + b + b; i++) num[i] = i - a - b;
  for (int i = a + b + b + 1; i <= a + a + b; i++)
    num[i] = a + (i - (a + b + b));
  for (int i = a + a + b + 1; i <= a + a + b + b; i++) num[i] = y;
  for (int i = 1; i <= 2 * (a + b); i++) num[i + 2 * (a + b)] = num[i];
  int ans = 0;
  if (r - l + 1 >= 2 * (a + b))
    ans = 2 * a - b;
  else {
    if (l == r) {
      ans = 1;
    } else {
      l = Get(l, 2 * (a + b));
      r = Get(r, 2 * (a + b));
      if (r <= l) r = r + 2 * (a + b);
      memset(cnt, 0, sizeof(cnt));
      for (int i = l; i <= r; i++) cnt[num[i]] = true;
      for (int i = 1; i <= 26; i++)
        if (cnt[i]) ans++;
    }
  }
  return ans;
}
void Solve2() {
  for (int i = 1; i <= a; i++) num[i] = i;
  for (int i = a + 1; i <= a + b; i++) num[i] = a;
  for (int i = a + b + 1; i <= a + b + a - 1; i++) num[i] = i - a - b;
  num[a + b + a] = a + 1;
  for (int i = a + a + b + 1; i <= a + a + b + b; i++) num[i] = a + 1;
  for (int i = 1; i <= 2 * (a + b); i++) num[i + 2 * (a + b)] = num[i];
  int ans = 0;
  if (r - l + 1 >= 2 * (a + b))
    ans = a + 1;
  else {
    if (l == r) {
      ans = 1;
    } else {
      l = Get(l, 2 * (a + b));
      r = Get(r, 2 * (a + b));
      if (r <= l) r = r + 2 * (a + b);
      memset(cnt, 0, sizeof(cnt));
      for (int i = l; i <= r; i++) cnt[num[i]] = true;
      for (int i = 1; i <= 26; i++)
        if (cnt[i]) ans++;
    }
  }
  printf("%d\n", ans);
}
int main() {
  while (~scanf("%d%d%d%d", &a, &b, &l, &r)) {
    if (a > b) {
      int ans = 1 << 30;
      for (int x = b + 1; x <= a; x++)
        for (int y = a + 1; y <= 2 * a - b; y++) ans = min(ans, Solve1(x, y));
      printf("%d\n", ans);
    } else
      Solve2();
  }
  return 0;
}
