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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long MOD = 1000000007;
const int INF = 1000000007;
const long long LINF = 1LL * INF * INF;
const int MAXN = 5047;
const double EPS = 1e-6;
const double PI = acos(-1.0);
int a, b, l, r, ans = INF;
void f(char c1, char c2) {
  string S;
  set<char> Se;
  for (int i = (0); i < (a); ++i) S.push_back('a' + i);
  for (int i = (0); i < (b); ++i) S.push_back(c1);
  for (int i = (0); i < (26); ++i) Se.insert('a' + i);
  for (int i = (b); i < (a + b); ++i) Se.erase(S[i]);
  for (int i = (0); i < (a); ++i) {
    S.push_back(*Se.begin());
    Se.erase(Se.begin());
  }
  for (int i = (0); i < (b); ++i) S.push_back(c2);
  for (int i = (0); i < (26); ++i) Se.insert('a' + i);
  for (int i = (a + 2 * b); i < (2 * (a + b)); ++i) Se.erase(S[i]);
  for (int i = (0); i < (a); ++i) {
    S.push_back(*Se.begin());
    Se.erase(Se.begin());
  }
  Se.clear();
  for (int i = (l); i < (r + 1); ++i) Se.insert(S[i]);
  ans = min(ans, (int)((Se).size()));
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> a >> b >> l >> r;
  --l, --r;
  if (b >= a && r >= (l / (a + b) + 1) * (a + b) + a - 1) {
    cout << a + 1;
    return 0;
  }
  if (b < a && r >= ((l - 1) / (a + b) + 2) * (a + b) + a - 1) {
    cout << 2 * a - b;
    return 0;
  }
  r -= l / (a + b) * (a + b);
  l %= a + b;
  for (int i = (0); i < (26); ++i)
    for (int j = (0); j < (26); ++j) f('a' + i, 'a' + j);
  cout << ans;
  return 0;
}
