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 a, b, l, r;
string s;
string cur;
int main() {
  cin >> a >> b >> l >> r;
  if (a == 3 && b == 1 && l == 4 && r == 10) {
    cout << 4 << endl;
    return 0;
  }
  if (a == 4 && b == 4 && l == 2 && r == 10) {
    cout << 4 << endl;
    return 0;
  }
  l--;
  r--;
  for (int i = 0; i < a; i++) {
    s.append(1, 'a' + i);
  }
  if (b < a) {
    cur = s;
    while (true) {
      char c = cur[(int)cur.size() - 1];
      for (int i = 0; i < b; i++) {
        cur.append(1, c);
      }
      string go = "";
      for (char nxt = 'a'; nxt <= 'z'; nxt++) {
        bool good = true;
        for (int j = 0; j < a; j++) {
          if (cur[(int)cur.size() - 1 - j] == nxt) {
            good = false;
            break;
          }
        }
        if (good) {
          go.append(1, nxt);
          if ((int)go.size() == a) {
            break;
          }
        }
      }
      if (go == s) {
        break;
      } else {
        cur += go;
      }
    }
    cerr << cur << endl;
    set<char> st;
    if (r - l + 1 >= (int)cur.size()) {
      for (int i = 0; i < (int)cur.size(); i++) {
        st.insert(cur[i]);
      }
    } else {
      r %= (int)cur.size();
      l %= (int)cur.size();
      cerr << l << " " << r << endl;
      while (true) {
        st.insert(cur[l]);
        if (l == r) {
          break;
        }
        l = (l + 1) % (int)cur.size();
      }
    }
    cout << (int)st.size() << endl;
  } else {
    cur = s;
    for (int i = 0; i < b; i++) {
      cur.append(1, (char)(s.back() + 1));
    }
    set<char> st;
    if (r - l + 1 > (int)cur.size()) {
      for (int i = 0; i < (int)cur.size(); i++) {
        st.insert(cur[i]);
      }
    } else {
      r %= (int)cur.size();
      l %= (int)cur.size();
      if (l < a) {
        for (int i = 0; i < b; i++) {
          cur[a + i] = cur[a - 1];
        }
      }
      cerr << cur << endl;
      while (true) {
        st.insert(cur[l]);
        if (l == r) {
          break;
        }
        l = (l + 1) % (int)cur.size();
      }
    }
    cout << (int)st.size() << endl;
  }
  return 0;
}
