#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
bool Max(T1& a, T2 b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T1, typename T2>
bool Min(T1& a, T2 b) {
  return a > b ? a = b, 1 : 0;
}
int arr[48];
void solve() {
  long long a, b, l, r;
  cin >> a >> b >> l >> r;
  for (int i = 0; i < (a); ++i) arr[i] = i;
  for (int i = a; i <= (a + b - 1); ++i) arr[i] = a - 1;
  long long t = min(b, a - 1);
  for (int i = 0; i < (t); ++i) arr[a + b + i] = i;
  for (int i = 0; i < (a - t); ++i) arr[a + b + t + i] = a + i;
  for (int i = 0; i < (b); ++i) arr[a + b + a + i] = a + a - t - 1;
  long long n = (a + b) * 2;
  l--;
  set<int> s;
  if (r - l >= n) {
    for (int i = 0; i < (n); ++i) s.insert(arr[i]);
  } else {
    l %= n;
    r %= n;
    if (b < a) {
      if (l >= a && l < a + b) {
        for (int i = a; i <= (a + b - 1); ++i) arr[i] = b;
      } else if (l >= a + b + a) {
        for (int i = 0; i < (b); ++i) arr[a + b + a + i] = a;
      }
    }
    for (; l != r; l = (l + 1) % n) s.insert(arr[l]);
  }
  cout << s.size() << '\n';
}
void init() {}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(20);
  init();
  solve();
  return 0;
}

### 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;
template <typename T1, typename T2>
bool Max(T1& a, T2 b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T1, typename T2>
bool Min(T1& a, T2 b) {
  return a > b ? a = b, 1 : 0;
}
int arr[48];
void solve() {
  long long a, b, l, r;
  cin >> a >> b >> l >> r;
  for (int i = 0; i < (a); ++i) arr[i] = i;
  for (int i = a; i <= (a + b - 1); ++i) arr[i] = a - 1;
  long long t = min(b, a - 1);
  for (int i = 0; i < (t); ++i) arr[a + b + i] = i;
  for (int i = 0; i < (a - t); ++i) arr[a + b + t + i] = a + i;
  for (int i = 0; i < (b); ++i) arr[a + b + a + i] = a + a - t - 1;
  long long n = (a + b) * 2;
  l--;
  set<int> s;
  if (r - l >= n) {
    for (int i = 0; i < (n); ++i) s.insert(arr[i]);
  } else {
    l %= n;
    r %= n;
    if (b < a) {
      if (l >= a && l < a + b) {
        for (int i = a; i <= (a + b - 1); ++i) arr[i] = b;
      } else if (l >= a + b + a) {
        for (int i = 0; i < (b); ++i) arr[a + b + a + i] = a;
      }
    }
    for (; l != r; l = (l + 1) % n) s.insert(arr[l]);
  }
  cout << s.size() << '\n';
}
void init() {}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(20);
  init();
  solve();
  return 0;
}

```