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;
long long l, r;
int op;
int dfs(int ch, vector<int> v) {
  if (v.size() >= 2 * (a + b)) {
    int cnt = 0;
    int vis[30] = {0};
    if (op == 1) {
      for (int i = 2 * (a + b) - a; i < 2 * (a + b); i++) {
        if (v[i] < a) return 0x3f3f3f3f;
      }
      for (int i = l; i < 2 * (a + b); i++) {
        if (vis[v[i]] == 0) cnt++;
        vis[v[i]] = 1;
      }
      for (int i = 0; i <= r; i++) {
        if (vis[v[i]] == 0) cnt++;
        vis[v[i]] = 1;
      }
    } else {
      for (int i = l; i <= r; i++) {
        if (vis[v[i]] == 0) cnt++;
        vis[v[i]] = 1;
      }
    }
    return cnt;
  }
  for (int i = 0; i < b; i++) v.push_back(ch);
  int vis[30] = {0};
  for (int i = v.size() - 1; i >= v.size() - a; i--) vis[v[i]] = 1;
  vector<int> box;
  for (int i = 0; i < 26; i++)
    if (vis[i] == 0) box.push_back(i);
  if (box.size() < a) return 0x3f3f3f3f;
  for (int i = 0; i < a; i++) {
    v.push_back(box[i]);
  }
  int ans = 0x3f3f3f3f;
  for (int i = 0; i < 26; i++) {
    ans = min(ans, dfs(i, v));
  }
  return ans;
}
int main() {
  scanf("%d%d%lld%lld", &a, &b, &l, &r);
  long long T = 2 * (a + b);
  int ans = 0x3f3f3f3f;
  if (r - l + 1 < T) {
    l--;
    r--;
    l %= T;
    r %= T;
    if (l > r) op = 1;
    vector<int> v;
    for (int i = 0; i < a; i++) v.push_back(i);
    for (int i = 0; i < 26; i++) {
      ans = min(ans, dfs(i, v));
    }
  } else {
    if (b >= a)
      ans = a + 1;
    else
      ans = 2 * a - b;
  }
  printf("%d\n", ans);
  return 0;
}
