#include <bits/stdc++.h>
using namespace std;
constexpr double pi = 3.141592653589793238462643383279502884L;
const int sz = 1e5;
bool used[50];
int main() {
  ios::sync_with_stdio(false);
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  --l, --r;
  if (l / (a + b) == r / (a + b)) {
    int lr = min(l % (a + b), a - 1);
    int rr = min(r % (a + b), a - 1);
    cout << rr - lr + 1 << endl;
  } else if (l / (a + b) + 1 == r / (a + b)) {
    int rc = min(r % (a + b), a - 1) + 1;
    int lc;
    if (a <= b) {
      lc = max(a - max(l % (a + b), rc), 1);
    } else {
      lc = max(a - max(min(rc, b), l % (a + b)), 1);
    }
    cout << rc + lc << endl;
  } else if (r - l + 1 >= 2 * (a + b)) {
    cout << a + max(a - b, 1) << endl;
  } else {
    assert(l / (a + b) + 2 == r / (a + b));
    int lc = max(a - max(b, min(l % (a + b), a)), 0);
    int rc = max(max(b - 1, min(r % (a + b), a)) - b + 1, 0);
    cout << rc + lc + a + (b >= a) << endl;
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
constexpr double pi = 3.141592653589793238462643383279502884L;
const int sz = 1e5;
bool used[50];
int main() {
  ios::sync_with_stdio(false);
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  --l, --r;
  if (l / (a + b) == r / (a + b)) {
    int lr = min(l % (a + b), a - 1);
    int rr = min(r % (a + b), a - 1);
    cout << rr - lr + 1 << endl;
  } else if (l / (a + b) + 1 == r / (a + b)) {
    int rc = min(r % (a + b), a - 1) + 1;
    int lc;
    if (a <= b) {
      lc = max(a - max(l % (a + b), rc), 1);
    } else {
      lc = max(a - max(min(rc, b), l % (a + b)), 1);
    }
    cout << rc + lc << endl;
  } else if (r - l + 1 >= 2 * (a + b)) {
    cout << a + max(a - b, 1) << endl;
  } else {
    assert(l / (a + b) + 2 == r / (a + b));
    int lc = max(a - max(b, min(l % (a + b), a)), 0);
    int rc = max(max(b - 1, min(r % (a + b), a)) - b + 1, 0);
    cout << rc + lc + a + (b >= a) << endl;
  }
}

```