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 <typename T>
inline void read(T& x) {
  cin >> x;
}
template <typename T, typename... Types>
inline void read(T& x, Types&... args) {
  cin >> x;
}
string build(long long a, long long b, char mr1, char mr2,
             long long reps = 16) {
  string s;
  for (long long i = 0; i < a; i++) s += 'a' + i;
  for (long long i = 0; i < reps; i++) {
    if (i & 1) {
      string suf = s.substr((long long)s.size() - a);
      set<char> st((suf).begin(), (suf).end());
      long long ptr = 0;
      for (long long j = 0; j < a; j++) {
        while (st.count('a' + ptr)) ptr++;
        st.insert('a' + ptr);
        s += 'a' + ptr;
      }
    } else {
      for (long long j = 0; j < b; j++) s += i % 4 == 0 ? mr1 : mr2;
    }
  }
  return s;
}
long long dumb(long long a, long long b, long long L, long long R, char mr1,
               char mr2) {
  string s = build(a, b, mr1, mr2, 100);
  assert(s.size() > R);
  set<char> st;
  for (long long i = L; i <= R; i++) {
    st.insert(s[i]);
  }
  return st.size();
}
long long solve(long long a, long long b, long long L, long long R, char mr1,
                char mr2) {
  string s = build(a, b, mr1, mr2);
  long long sz = (long long)s.size() - a;
  set<char> st;
  for (long long i = 0; i < a; i++)
    if (i >= L && i <= R) st.insert(s[i]);
  string ss = s.substr(a);
  L = max(L, a);
  if (R - L + 1 > 4LL * ss.size()) {
    for (char c : ss) st.insert(c);
  } else {
    for (long long i = L; i <= R; i++) {
      st.insert(ss[(i - a) % sz]);
    }
  }
  return st.size();
}
int32_t main() {
  long long a, b, L, R;
  cin >> a >> b >> L >> R;
  L--, R--;
  long long ans = 1e18;
  for (long long i = 0; i < 26; i++) {
    for (long long j = 0; j < 26; j++) {
      ans = min(ans, solve(a, b, L, R, 'a' + i, 'a' + j));
    }
  }
  cout << ans << endl;
}
