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() {
  ios::sync_with_stdio(false);
  int a, b, l, r, i;
  cin >> a >> b >> l >> r;
  string d = "", s = "";
  for (i = 0; i < a; i++) s = s + (char)('a' + i);
  d = s;
  for (i = 0; i < b; i++) {
    s = s + (char)('a' + a - 1);
    d = d + (char)('a' + b);
  }
  if (a <= b + 1) {
    for (i = 0; i < a - 1; i++) s = s + (char)('a' + i);
    s = s + (char)('a' + a);
  } else {
    for (i = 0; i < b; i++) s = s + (char)('a' + i);
    for (i = 0; i < a - b; i++) s = s + (char)('a' + a + i);
  }
  char h = s[s.size() - 1];
  for (i = 0; i < b; i++) s = s + h;
  int n = 2 * (a + b);
  s = s + s;
  s = s + s;
  if (l == r) {
    cout << 1;
    return 0;
  }
  bool t = false;
  if (r - l >= n) {
    l = 0;
    r = n - 1;
  } else {
    l--;
    r--;
    l = l % n;
    r = r % n;
    if (r < l) r += n;
  }
  if (a > b) {
    bool arr[26];
    memset(arr, 0, sizeof arr);
    for (i = b + a - 1; i >= b; i--) arr[d[i] - 'a'] = 1;
    i = 0;
    while (d.size() != a + a + b) {
      if (!arr[i]) d = d + (char)('a' + i);
      i++;
      if (d.size() <= a + b + b) arr[d[i - 1] - 'a'] = 1;
    }
    for (i = 0; i < 26; i++)
      if (!arr[i]) break;
    h = (char)('a' + i);
    for (i = 0; i < b; i++) d = d + h;
    d = d + d;
  }
  set<char> k;
  set<char> g;
  for (i = l; i <= r; i++) {
    k.insert(s[i]);
    if (a > b) g.insert(d[i]);
  }
  if (a > b)
    cout << (k.size() < g.size() ? k.size() : g.size());
  else
    cout << k.size();
  return 0;
}
