#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  bool loops_around = (l - 1) % (a + b) >= a && (r - 1) % (a + b) < a;
  vector<string> first_four(4);
  for (int i = 0; i < a; ++i) {
    first_four[0].push_back('a' + i);
  }
  for (int i = 0; i < b; ++i) {
    if (a > b && loops_around) {
      first_four[1].push_back(first_four[0][b]);
    } else {
      first_four[1].push_back(first_four[0].back());
    }
  }
  if (a <= b) {
    for (int i = 0; i < a - 1; ++i) {
      first_four[2].push_back('a' + i);
    }
    first_four[2].push_back(first_four[1].back() + 1);
  } else {
    for (int i = 0; i < b; ++i) {
      first_four[2].push_back('a' + i);
    }
    for (int i = 0; i < a - b; ++i) {
      first_four[2].push_back(first_four[1].back() + i + 1);
    }
  }
  for (int i = 0; i < b; ++i) {
    if (a > b && loops_around) {
      first_four[3].push_back(first_four[2][b]);
    } else {
      first_four[3].push_back(first_four[2].back());
    }
  }
  string ans = first_four[0] + first_four[1] + first_four[2] + first_four[3];
  unordered_set<char> seen;
  for (int i = l; i <= min(r, l + (int)ans.size() - 1); ++i) {
    int pos = (i - 1) % ans.size();
    seen.insert(ans[pos]);
  }
  cout << seen.size() << "\n";
  return 0;
}

### Prompt

Develop a solution in CPP 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  bool loops_around = (l - 1) % (a + b) >= a && (r - 1) % (a + b) < a;
  vector<string> first_four(4);
  for (int i = 0; i < a; ++i) {
    first_four[0].push_back('a' + i);
  }
  for (int i = 0; i < b; ++i) {
    if (a > b && loops_around) {
      first_four[1].push_back(first_four[0][b]);
    } else {
      first_four[1].push_back(first_four[0].back());
    }
  }
  if (a <= b) {
    for (int i = 0; i < a - 1; ++i) {
      first_four[2].push_back('a' + i);
    }
    first_four[2].push_back(first_four[1].back() + 1);
  } else {
    for (int i = 0; i < b; ++i) {
      first_four[2].push_back('a' + i);
    }
    for (int i = 0; i < a - b; ++i) {
      first_four[2].push_back(first_four[1].back() + i + 1);
    }
  }
  for (int i = 0; i < b; ++i) {
    if (a > b && loops_around) {
      first_four[3].push_back(first_four[2][b]);
    } else {
      first_four[3].push_back(first_four[2].back());
    }
  }
  string ans = first_four[0] + first_four[1] + first_four[2] + first_four[3];
  unordered_set<char> seen;
  for (int i = l; i <= min(r, l + (int)ans.size() - 1); ++i) {
    int pos = (i - 1) % ans.size();
    seen.insert(ans[pos]);
  }
  cout << seen.size() << "\n";
  return 0;
}

```