#include <bits/stdc++.h>
using namespace std;
const int maxn = 5100;
const int INF = 1 << 30;
const int MOD = 1e9 + 7;
const double dif = 1e-5;
const double PI = 3.14159265358979323846;
int n, m;
int main(void) {
  int a, b, l, r, t1 = -1, t2 = -1, t3 = -1;
  while (cin >> a >> b >> l >> r) {
    while ((r % (a + b) > a) || (r % (a + b) == 0)) r--;
    if (r <= l)
      cout << 1 << endl;
    else if ((t1 = min(r - l + 1, (r - (r / (a + b)) * (a + b)))) ==
             (r - l + 1))
      cout << t1 << endl;
    else if ((t2 = min(r - l + 1, t1 + b)) == (r - l + 1))
      cout << t1 + 1 << endl;
    else if ((t3 = (r - t2 - l + 1)) > a && b >= a)
      cout << a + 1 << endl;
    else if (b >= a)
      cout << t1 + 1 + max(0, min(t3, a - t1 - 1)) << endl;
    else if (t3 <= (a - b))
      cout << t1 + t3 << endl;
    else if (t3 <= a)
      cout << t3 + min(t1, a - t3) + max(0, t1 - b) << endl;
    else if (t3 <= (a + b))
      cout << a + 1 + max(0, min(t1 - b - 1, a - b - 1)) << endl;
    else
      cout << a + max(0, t1 - b) + min(t3 - (a + b), a - b - max(0, t1 - b))
           << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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;
const int maxn = 5100;
const int INF = 1 << 30;
const int MOD = 1e9 + 7;
const double dif = 1e-5;
const double PI = 3.14159265358979323846;
int n, m;
int main(void) {
  int a, b, l, r, t1 = -1, t2 = -1, t3 = -1;
  while (cin >> a >> b >> l >> r) {
    while ((r % (a + b) > a) || (r % (a + b) == 0)) r--;
    if (r <= l)
      cout << 1 << endl;
    else if ((t1 = min(r - l + 1, (r - (r / (a + b)) * (a + b)))) ==
             (r - l + 1))
      cout << t1 << endl;
    else if ((t2 = min(r - l + 1, t1 + b)) == (r - l + 1))
      cout << t1 + 1 << endl;
    else if ((t3 = (r - t2 - l + 1)) > a && b >= a)
      cout << a + 1 << endl;
    else if (b >= a)
      cout << t1 + 1 + max(0, min(t3, a - t1 - 1)) << endl;
    else if (t3 <= (a - b))
      cout << t1 + t3 << endl;
    else if (t3 <= a)
      cout << t3 + min(t1, a - t3) + max(0, t1 - b) << endl;
    else if (t3 <= (a + b))
      cout << a + 1 + max(0, min(t1 - b - 1, a - b - 1)) << endl;
    else
      cout << a + max(0, t1 - b) + min(t3 - (a + b), a - b - max(0, t1 - b))
           << endl;
  }
  return 0;
}

```