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 static int inf = 1e9 + 7;
const static int N = 1e4 + 5;
char x[65];
int c[26];
vector<int> d;
int main() {
  int a, b, l, r;
  scanf("%d%d%d%d", &a, &b, &l, &r);
  for (int i = 0; i < 26; ++i) c[i] = 0;
  for (int i = 0; i < a; ++i) x[i] = 'a' + i;
  for (int i = a; i < a + b; ++i) x[i] = x[i - 1];
  for (int i = 0; i < 26; ++i) c[i] = 0;
  for (int i = a + b - 1; i >= b; --i) c[x[i] - 'a']++;
  int y = (l - 1) % (a + b);
  for (int i = a + b; i < a + b + a; ++i)
    for (int j = 0; j < 26; ++j)
      if (c[j] == 0) {
        x[i] = 'a' + j;
        c[j] = 1;
        break;
      }
  for (int i = a + b + a; i < a + b + a + b; ++i) x[i] = x[i - 1];
  for (int i = 0; i < 26; ++i) c[i] = 0;
  for (int i = a + b + a + b - 1; i >= a + b + b; --i) c[x[i] - 'a']++;
  d.clear();
  for (int i = y; i < a + b + a - 1; ++i) d.push_back(x[i] - 'a');
  int w = 0;
  for (int i = 0; i < d.size(); ++i)
    if (c[d[i]] == 0) {
      x[a + b + a + b + w] = 'a' + d[i];
      c[d[i]]++;
      w++;
    }
  for (int i = a + b + a + b + w; i < a + b + a + b + a; ++i)
    for (int j = 0; j < 26; ++j)
      if (c[j] == 0) {
        x[i] = 'a' + j;
        c[j] = 1;
        break;
      }
  int z = r - l;
  for (int i = 0; i < 26; ++i) c[i] = 0;
  for (int i = y; i < a + b + a + b + a && i <= z + y; ++i) c[x[i] - 'a']++;
  int sum = 0;
  for (int i = 0; i < 26; ++i)
    if (c[i] != 0) sum++;
  printf("%d\n", sum);
  return 0;
}
