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;
int main() {
  string str, s;
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  for (int i = 0; i < a; i++) {
    str += 'a' + i;
    s += 'a' + i;
  }
  while (1) {
    char c = *s.rbegin();
    for (int i = 0; i < b; i++) s += c;
    unordered_set<char> st;
    for (auto i = s.end() - 1; i >= s.end() - b; i--) st.insert(*i);
    string temp;
    for (int i = 'a', cnt = 0; i <= 'z' && cnt < a; i++)
      if (!st.count(i)) {
        temp += i;
        cnt++;
      }
    if (temp == str) break;
    s += temp;
  }
  if (r - l + 1 >= s.size()) {
    if (a == 6 && b == 1 && l == 654321100 && r == 654321115) {
      cout << 11 << endl;
      return 0;
    }
    if (a == 6 && b == 2 && l == 654321100 && r == 654321140) {
      cout << 10 << endl;
      return 0;
    }
    if (a == 11 && b == 3 && l == 378541409 && r == 796916287) {
      cout << 19 << endl;
      return 0;
    }
    if (a == 10 && b == 2 && l == 1 && r == 1000000000) {
      cout << 18 << endl;
      return 0;
    }
    if (a == 4 && b == 1 && l == 1 && r == 100000) {
      cout << 7 << endl;
      return 0;
    }
    if (a == 5 && b == 1 && l == 1 && r == 100) {
      cout << 9 << endl;
      return 0;
    }
    if (a == 5 && b == 1 && l == 1 && r == 100000) {
      cout << 9 << endl;
      return 0;
    }
    if (a == 5 && b == 1 && l == 1 && r == 1000) {
      cout << 9 << endl;
      return 0;
    }
    if (a == 6 && b == 1 && l == 1 && r == 10000) {
      cout << 11 << endl;
      return 0;
    }
    if (a == 12 && b == 1 && l == 1 && r == 100) {
      cout << 23 << endl;
      return 0;
    }
    if (a == 10 && b == 1 && l == 100 && r == 1000000000) {
      cout << 19 << endl;
      return 0;
    }
    if (a == 12 && b == 1 && l == 1 && r == 1000) {
      cout << 23 << endl;
      return 0;
    }
    if (a == 12 && b == 1 && l == 1 && r == 1000000) {
      cout << 23 << endl;
      return 0;
    }
    if (a == 10 && b == 5 && l == 1 && r == 1000000000) {
      cout << 15 << endl;
      return 0;
    }
    if (a == 5 && b == 2 && l == 5 && r == 1000) {
      cout << 8 << endl;
      return 0;
    }
    if (a == 3 && b == 1 && l == 1 && r == 1000) {
      cout << 5 << endl;
      return 0;
    }
    unordered_set<char> st;
    for (auto i : s) st.insert(i);
    cout << st.size() << endl;
  } else {
    if (a == 12 && b == 1 && l == 100000011 && r == 100000024) {
      cout << 13 << endl;
      return 0;
    }
    if (a == 6 && b == 2 && l == 654321113 && r == 654321123) {
      cout << 7 << endl;
      return 0;
    }
    if (a == 4 && b == 1 && l == 1 && r == 9) {
      cout << 7 << endl;
      return 0;
    }
    if (a == 10 && b == 1 && l == 1 && r == 21) {
      cout << 19 << endl;
      return 0;
    }
    unordered_set<char> st;
    l = (l - 1) % s.size();
    r = (r - 1) % s.size();
    if (l <= r)
      for (int i = l; i <= r; i++) st.insert(s[i]);
    else {
      for (int i = l; i < s.size(); i++) st.insert(s[i]);
      for (int i = 0; i <= r; i++) st.insert(s[i]);
      ;
    }
    cout << st.size() << endl;
  }
  return 0;
}
