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;
void addmod(int &a, long long b) {
  a = (a + b);
  if (a >= 1000000007) a -= 1000000007;
}
void mulmod(int &a, long long b) { a = (a * b) % 1000000007; }
template <class T>
bool domin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
bool domax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
int gi() {
  int a;
  scanf("%d", &a);
  return a;
}
long long gll() {
  long long a;
  scanf("%lld", &a);
  return a;
}
void populate(string &st, set<char> &sc, int l, int r, int a, int b, int bi) {
  cerr << l << " " << r << " " << bi << endl;
  vector<bool> tkn(26, false);
  if (l && l % (a + b) == 0) {
    for (int i = max(0, bi - a); i < bi; i++)
      if (st[i] != ' ') {
        tkn[st[i] - 'a'] = true;
      }
  }
  int ix = 0;
  for (int i = bi, ti = l; ti <= r; i++, ti++) {
    if (l < a)
      st[i] = 'a' + ti, sc.insert('a' + ti);
    else {
      while (tkn[ix]) ix++;
      st[i] = ('a' + ix);
      sc.insert('a' + ix);
      ix++;
    }
  }
}
int main() {
  int a = gi(), b = gi(), l = gi() - 1, r = gi() - 1;
  int rl = r - l + 1;
  if (rl > 2 * (a + b)) {
    return cout << a + max(1, a - b), 0;
  }
  string sans(rl, ' ');
  set<char> sc;
  int lm = (l % (a + b)), si = 0, li = l;
  if (lm > 0) {
    if (lm < a) {
      populate(sans, sc, lm, a - 1, a, b, 0);
    }
    li += (a + b) - lm;
    si += (a + b) - lm;
  }
  for (; li <= r; li += (a + b), si += (a + b)) {
    populate(sans, sc, li, min(r, li + a - 1), a, b, si);
  }
  int ans = sc.size(), inc = 0;
  for (int ix = 0; ix < rl; ix++)
    if (sans[ix] == ' ') {
      set<char> ts;
      for (int j = ix + 1; j < rl && j < ix + a + 1; j++)
        if (sans[j] != ' ') {
          ts.insert(sans[j]);
        }
      if (ts.size() == sc.size()) inc = 1;
    }
  cout << ans + inc << endl;
  return 0;
}
