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 A, B, L, R;
int ov;
int ans = 1e9;
bool vis[30];
int p[50];
int mkpat(int x, int y) {
  int n = (A + B) * 2;
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i <= n - 1; i++) {
    if (i < A) {
      p[i] = i;
      if (i >= A - ov) vis[i] = true;
    } else if (i < A + B) {
      vis[x] = true;
      p[i] = x;
    } else if (i < A + B + A) {
      p[i] = i - (A + B);
      while (vis[p[i]]) p[i]++;
      vis[p[i]] = true;
    } else
      p[i] = y;
  }
}
int brute(int x, int y) {
  mkpat(x, y);
  int n = (A + B) * 2;
  set<int> S;
  for (int i = L; i <= R; i++) {
    if (i != L && i % n == L % n) break;
    int ci = p[i % n];
    S.insert(ci);
  }
  return S.size();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> A >> B >> L >> R;
  int n = (A + B) * 2, hn = (A + B);
  ov = max(0, (A - B));
  L--, R--;
  int ln = (R - L + 1);
  if (B >= A) {
    for (int i = 0; i <= n - 1; i++) {
      if (i < A)
        p[i] = i;
      else if (i < A + B)
        p[i] = A - 1;
      else if (i < A + B + A) {
        p[i] = i - (A + B);
        if (p[i] == A - 1) p[i]++;
      } else
        p[i] = A;
    }
    set<int> S[2];
    for (int i = L; i <= R; i++) {
      if (i != L && i % n == L % n) break;
      S[0].insert(p[i % n]);
      S[1].insert(p[(i + hn) % n]);
    }
    cout << min(S[0].size(), S[1].size()) << '\n';
    return 0;
  }
  if (ln >= n) {
    cout << A + ov << '\n';
    return 0;
  }
  for (int i = 0; i <= A - 1; i++)
    for (int j = A; j <= A + ov - 1; j++) ans = min(ans, brute(i, j));
  cout << ans << '\n';
  return 0;
}
