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.
#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << '{';
  for (int i = (0), __i = (((int)(vec).size())); i < __i; i++) {
    os << vec[i];
    if (i + 1 != ((int)(vec).size())) os << ',';
  }
  os << '}';
  return os;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &par) {
  os << '(' << par.first << ',' << par.second << ')';
  return os;
}
int gcd(int x, int y) { return y ? gcd(y, x % y) : abs(x); }
template <class T>
T sqr(T x) {
  return x * x;
}
bool used[26];
int main() {
  ios::sync_with_stdio(false);
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  int cycle = (a + b) * 2;
  int rest1 = (l - 1) % cycle;
  int rest2 = r % cycle;
  if (rest2 == 0) rest2 = cycle;
  int cnt = (r - 1) / cycle - (l - 1) / cycle;
  if (cnt > 1) {
    if (a <= b)
      cout << a + 1;
    else
      cout << 2 * a - b;
    return 0;
  }
  int xl = min(b, a - 1);
  int yl = a - xl;
  int zl = yl;
  string x(xl, ' '), y(yl, ' '), z(zl, ' ');
  for (int i = (0), __i = (xl); i < __i; i++) x[i] = 'a' + i;
  for (int i = (0), __i = (yl); i < __i; i++) y[i] = 'a' + xl + i;
  for (int i = (0), __i = (zl); i < __i; i++) z[i] = 'a' + xl + yl + i;
  string _x(x), _y(y), _z(z);
  reverse((_x).begin(), (_x).end());
  reverse((_y).begin(), (_y).end());
  reverse((_z).begin(), (_z).end());
  string s = x + y + string(b, ' ') + x + z + string(b, ' ');
  s += s;
  rest2 += cnt * cycle;
  for (int i = (rest1), __i = (rest2); i < __i; i++)
    if (s[i] != ' ') used[s[i] - 'a'] = true;
  for (int i = (0), __i = (rest2); i < __i; i++) {
    if (s[i] == ' ') {
      int j = i - 1;
      while (j >= i - yl) {
        if (used[s[j] - 'a']) {
          s[i] = s[j];
          break;
        }
        j--;
      }
      if (j < i - yl) {
        s[i] = s[i - 1];
        used[s[i] - 'a'] = true;
      }
    }
  }
  for (int i = (0), __i = (26); i < __i; i++) used[i] = false;
  for (int i = (rest1), __i = (rest2); i < __i; i++)
    if (s[i] != ' ') used[s[i] - 'a'] = true;
  int res = 0;
  for (int i = (0), __i = (26); i < __i; i++) res += used[i];
  cout << res;
  return 0;
}
