#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b, l, r, count = 0, i = 0, j, ans = 0, total, start;
  cin >> a;
  cin >> b;
  cin >> l;
  cin >> r;
  start = (l - 1) % (a + b);
  bool flag = false;
  char curr;
  total = 2 * a + 2 * b;
  unordered_map<char, int> m;
  string s = "";
  string temp = "1";
  while (i < total) {
    count = 0;
    if (flag) {
      m.clear();
      if (a <= b || start < a) {
        curr = s[i - 1];
        temp[0] = curr;
      } else {
        curr = s[i - (a - b)];
        temp[0] = curr;
      }
      while (count < b) {
        s += temp;
        i++;
        count++;
      }
      j = i - 1;
      count = 0;
      while (count < a) {
        m[s[j]]++;
        count++;
        j--;
      }
      flag = !flag;
    } else {
      curr = 'a';
      while (count < a) {
        if (m.find(curr) == m.end()) {
          temp[0] = curr;
          s += temp;
          curr++;
          count++;
          i++;
        } else {
          curr++;
        }
      }
      flag = !flag;
    }
  }
  m.clear();
  if ((r - l + 1) >= total) {
    for (i = 0; i < total; i++) {
      if (m.find(s[i]) == m.end()) {
        ans++;
        m[s[i]]++;
      }
    }
  } else {
    l--;
    r--;
    l = l % total;
    r = r % total;
    if (l <= r) {
      for (i = l; i <= r; i++) {
        if (m.find(s[i]) == m.end()) {
          ans++;
          m[s[i]]++;
        }
      }
    } else {
      for (i = l; i < total; i++) {
        if (m.find(s[i]) == m.end()) {
          ans++;
          m[s[i]]++;
        }
      }
      for (i = 0; i <= r; i++) {
        if (m.find(s[i]) == m.end()) {
          ans++;
          m[s[i]]++;
        }
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Please formulate 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;
int main() {
  int a, b, l, r, count = 0, i = 0, j, ans = 0, total, start;
  cin >> a;
  cin >> b;
  cin >> l;
  cin >> r;
  start = (l - 1) % (a + b);
  bool flag = false;
  char curr;
  total = 2 * a + 2 * b;
  unordered_map<char, int> m;
  string s = "";
  string temp = "1";
  while (i < total) {
    count = 0;
    if (flag) {
      m.clear();
      if (a <= b || start < a) {
        curr = s[i - 1];
        temp[0] = curr;
      } else {
        curr = s[i - (a - b)];
        temp[0] = curr;
      }
      while (count < b) {
        s += temp;
        i++;
        count++;
      }
      j = i - 1;
      count = 0;
      while (count < a) {
        m[s[j]]++;
        count++;
        j--;
      }
      flag = !flag;
    } else {
      curr = 'a';
      while (count < a) {
        if (m.find(curr) == m.end()) {
          temp[0] = curr;
          s += temp;
          curr++;
          count++;
          i++;
        } else {
          curr++;
        }
      }
      flag = !flag;
    }
  }
  m.clear();
  if ((r - l + 1) >= total) {
    for (i = 0; i < total; i++) {
      if (m.find(s[i]) == m.end()) {
        ans++;
        m[s[i]]++;
      }
    }
  } else {
    l--;
    r--;
    l = l % total;
    r = r % total;
    if (l <= r) {
      for (i = l; i <= r; i++) {
        if (m.find(s[i]) == m.end()) {
          ans++;
          m[s[i]]++;
        }
      }
    } else {
      for (i = l; i < total; i++) {
        if (m.find(s[i]) == m.end()) {
          ans++;
          m[s[i]]++;
        }
      }
      for (i = 0; i <= r; i++) {
        if (m.find(s[i]) == m.end()) {
          ans++;
          m[s[i]]++;
        }
      }
    }
  }
  cout << ans;
  return 0;
}

```