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;
string s;
string s1, s2;
string tmp;
map<pair<string, string>, int> mapa;
int a, b, l, r;
int find(int x) {
  x -= (int)s.size();
  x %= (int)tmp.size();
  return x;
}
int main() {
  scanf("%d %d %d %d", &a, &b, &l, &r);
  int sol = (1 << 30);
  int L = l;
  int R = r;
  for (int ttt = (0); ttt < int(100); ttt++) {
    mapa.clear();
    s = "";
    l = L;
    r = R;
    for (int i = ('a'); i < int('a' + a); i++) s += (char)i;
    int it = -1;
    while ((int)s.size() < r) {
      char c = s[(int)s.size() - (rand() % 2) - 1];
      s1 = s2 = "";
      for (int i = (0); i < int(b); i++) s1 += c;
      string t = s + s1;
      string t1 = "";
      for (int i = (0); i < int(a); i++) t1 += t[(int)t.size() - 1 - i];
      reverse((t1).begin(), (t1).end());
      for (int i = ('a'); i < int('z' + 1); i++) {
        bool ok = 1;
        for (int j = (0); j < int((int)t1.size()); j++)
          if (t1[j] == i) ok = 0;
        if (ok) s2 += (char)i;
        if ((int)s2.size() == a) break;
      }
      if (mapa.find(make_pair(s1, s2)) != mapa.end()) {
        it = mapa[make_pair(s1, s2)];
        break;
      }
      mapa[make_pair(s1, s2)] = (int)s.size();
      s += s1 + s2;
    }
    if ((int)s.size() >= r) {
      l--;
      r--;
      set<char> st;
      for (int i = (l); i < int(r + 1); i++) st.insert(s[i]);
      printf("%d\n", (int)st.size());
      return 0;
    }
    tmp = s1 + s2;
    set<char> st;
    if (l - 1 < (int)s.size()) {
      l--;
      for (int i = (l); i < int((int)s.size()); i++) st.insert(s[i]);
      l = (int)s.size() + 1;
    }
    tmp = s.substr(it, (int)s.size() - it);
    l--;
    r--;
    l -= (int)s.size();
    r -= (int)s.size();
    if (r - l + 1 >= (int)tmp.size())
      for (int i = (0); i < int((int)tmp.size()); i++) st.insert(tmp[i]);
    else {
      string kraj = tmp + tmp;
      int x = l % (int)tmp.size();
      int y = r % (int)tmp.size();
      if (x > y) y += (int)tmp.size();
      for (int i = (x); i < int(y + 1); i++) st.insert(kraj[i]);
    }
    sol = min(sol, (int)st.size());
  }
  printf("%d\n", sol);
  return 0;
}
