#include <bits/stdc++.h>
using namespace std;
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  int count;
  if (b >= a) {
    count = 0;
    map<int, int> tca;
    int v[4 * (a + b)];
    int last;
    for (long long int i = (0); i <= (a - 1); i++) v[i] = i;
    for (long long int i = (a); i <= (a + b - 1); i++) v[i] = v[a - 1];
    last = v[a - 1];
    for (long long int i = (a + b); i <= (2 * a + b - 2); i++)
      v[i] = v[i - (a + b)];
    for (long long int i = (2 * a + b - 1); i <= (2 * a + 2 * b - 1); i++)
      v[i] = last + 1;
    l--, r--;
    int k = l % (2 * (a + b));
    r = r - (l - k);
    for (long long int i = (2 * (a + b)); i <= (4 * (a + b) - 1); i++)
      v[i] = v[i - 2 * (a + b)];
    while (k <= r && k < 4 * (a + b)) {
      if (tca[v[k]] == 0) count++, tca[v[k]] = 1;
      k++;
    }
  } else {
    count = 0;
    map<int, int> tca;
    int v[4 * (a + b)];
    int last;
    for (long long int i = (0); i <= (a - 1); i++) v[i] = i;
    for (long long int i = (a); i <= (a + b - 1); i++) v[i] = v[a - 1];
    last = v[a - 1];
    for (long long int i = (a + b); i <= (a + b + b - 1); i++)
      v[i] = v[i - (a + b)];
    for (long long int i = (a + 2 * b); i <= (2 * a + b - 1); i++)
      v[i] = (++last);
    for (long long int i = (2 * a + b); i <= (2 * a + 2 * b - 1); i++)
      v[i] = v[2 * a + b - 1];
    l--, r--;
    int k = l % (2 * (a + b)), k2 = l % (2 * (a + b));
    r = r - (l - k);
    for (long long int i = (2 * (a + b)); i <= (4 * (a + b) - 1); i++)
      v[i] = v[i - 2 * (a + b)];
    while (k <= r && k < 4 * (a + b)) {
      if (tca[v[k]] == 0) count++, tca[v[k]] = 1;
      k++;
    }
    if ((a <= k2 && k2 <= a + b - 1 && r == 3 * a + 2 * b - 2) ||
        (2 * a + b <= k2 && k2 <= 2 * a + 2 * b - 1 && r == 4 * a + 3 * b - 2))
      count--;
  }
  cout << count << "\n";
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  int count;
  if (b >= a) {
    count = 0;
    map<int, int> tca;
    int v[4 * (a + b)];
    int last;
    for (long long int i = (0); i <= (a - 1); i++) v[i] = i;
    for (long long int i = (a); i <= (a + b - 1); i++) v[i] = v[a - 1];
    last = v[a - 1];
    for (long long int i = (a + b); i <= (2 * a + b - 2); i++)
      v[i] = v[i - (a + b)];
    for (long long int i = (2 * a + b - 1); i <= (2 * a + 2 * b - 1); i++)
      v[i] = last + 1;
    l--, r--;
    int k = l % (2 * (a + b));
    r = r - (l - k);
    for (long long int i = (2 * (a + b)); i <= (4 * (a + b) - 1); i++)
      v[i] = v[i - 2 * (a + b)];
    while (k <= r && k < 4 * (a + b)) {
      if (tca[v[k]] == 0) count++, tca[v[k]] = 1;
      k++;
    }
  } else {
    count = 0;
    map<int, int> tca;
    int v[4 * (a + b)];
    int last;
    for (long long int i = (0); i <= (a - 1); i++) v[i] = i;
    for (long long int i = (a); i <= (a + b - 1); i++) v[i] = v[a - 1];
    last = v[a - 1];
    for (long long int i = (a + b); i <= (a + b + b - 1); i++)
      v[i] = v[i - (a + b)];
    for (long long int i = (a + 2 * b); i <= (2 * a + b - 1); i++)
      v[i] = (++last);
    for (long long int i = (2 * a + b); i <= (2 * a + 2 * b - 1); i++)
      v[i] = v[2 * a + b - 1];
    l--, r--;
    int k = l % (2 * (a + b)), k2 = l % (2 * (a + b));
    r = r - (l - k);
    for (long long int i = (2 * (a + b)); i <= (4 * (a + b) - 1); i++)
      v[i] = v[i - 2 * (a + b)];
    while (k <= r && k < 4 * (a + b)) {
      if (tca[v[k]] == 0) count++, tca[v[k]] = 1;
      k++;
    }
    if ((a <= k2 && k2 <= a + b - 1 && r == 3 * a + 2 * b - 2) ||
        (2 * a + b <= k2 && k2 <= 2 * a + 2 * b - 1 && r == 4 * a + 3 * b - 2))
      count--;
  }
  cout << count << "\n";
  return 0;
}

```