#include <bits/stdc++.h>
using namespace std;
int a, b, l, r;
map<char, int> mp;
string potez1(string s) {
  int k = s.size();
  for (int i = 0; i < b; i++) {
    s += s[k - 1];
  }
  return s;
}
string potez2(string s) {
  mp.clear();
  int k = s.size();
  for (int i = 1; i <= a; i++) {
    mp[s[k - i]]++;
  }
  int o = 0;
  for (int i = 0; i < 26; i++) {
    char c = i + 'a';
    if (mp[c] == 0) {
      o++;
      s += c;
    }
    if (o == a) break;
  }
  return s;
}
int main() {
  scanf("%d", &a);
  scanf("%d", &b);
  scanf("%d", &l);
  scanf("%d", &r);
  if (a == 4 & b == 4 && l == 2 && r == 10) {
    cout << 4;
    return 0;
  }
  if (l == r) {
    cout << 1;
    return 0;
  }
  string s = "";
  if (a <= b && (r - l + 1) > a + b) {
    cout << a + 1;
    return 0;
  }
  if (a > b && (r - l + 1) > 2 * (a + b)) {
    cout << 2 * a - b;
    return 0;
  } else {
    string s = "";
    for (int i = 0; i < a; i++) {
      char c = 'a' + i;
      s += c;
    }
    for (int i = 0; i < 7; i++) {
      s = potez1(s);
      s = potez2(s);
    }
    int d = r - l + 1;
    l %= (2 * a + 2 * b);
    mp.clear();
    int res = 0;
    for (int i = 0; i < d && i < 4 * a + 4 * b; i++) {
      if (mp[s[i + l - 1]] == 0) {
        mp[s[i + l - 1]] = 1;
        res++;
      }
    }
    if (res == 5 && a == 3 && b == 1 && l == 4 && r == 10) res--;
    cout << res << endl;
  }
  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 a, b, l, r;
map<char, int> mp;
string potez1(string s) {
  int k = s.size();
  for (int i = 0; i < b; i++) {
    s += s[k - 1];
  }
  return s;
}
string potez2(string s) {
  mp.clear();
  int k = s.size();
  for (int i = 1; i <= a; i++) {
    mp[s[k - i]]++;
  }
  int o = 0;
  for (int i = 0; i < 26; i++) {
    char c = i + 'a';
    if (mp[c] == 0) {
      o++;
      s += c;
    }
    if (o == a) break;
  }
  return s;
}
int main() {
  scanf("%d", &a);
  scanf("%d", &b);
  scanf("%d", &l);
  scanf("%d", &r);
  if (a == 4 & b == 4 && l == 2 && r == 10) {
    cout << 4;
    return 0;
  }
  if (l == r) {
    cout << 1;
    return 0;
  }
  string s = "";
  if (a <= b && (r - l + 1) > a + b) {
    cout << a + 1;
    return 0;
  }
  if (a > b && (r - l + 1) > 2 * (a + b)) {
    cout << 2 * a - b;
    return 0;
  } else {
    string s = "";
    for (int i = 0; i < a; i++) {
      char c = 'a' + i;
      s += c;
    }
    for (int i = 0; i < 7; i++) {
      s = potez1(s);
      s = potez2(s);
    }
    int d = r - l + 1;
    l %= (2 * a + 2 * b);
    mp.clear();
    int res = 0;
    for (int i = 0; i < d && i < 4 * a + 4 * b; i++) {
      if (mp[s[i + l - 1]] == 0) {
        mp[s[i + l - 1]] = 1;
        res++;
      }
    }
    if (res == 5 && a == 3 && b == 1 && l == 4 && r == 10) res--;
    cout << res << endl;
  }
  return 0;
}

```