#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  l--, r--;
  if (r - l > 100) {
    cout << 2 * a - min(b, a - 1);
    return 0;
  }
  string s = "";
  for (int i = 0; i < a; i++) s += i + 'a';
  for (int i = 0; i < b; i++) s += s[a - 1];
  set<char> vis;
  for (int i = 1; i <= a; i++) vis.insert(s[int(s.size()) - i]);
  for (int i = 0, c = 0; i < 26 && c < a; i++) {
    if (vis.count('a' + i)) continue;
    s += 'a' + i;
    c++;
  }
  for (int i = 0; i < b; i++) s += s.back();
  string s2 = "";
  int ml = l % (2 * (a + b));
  while (r-- >= l) {
    s2 += s[ml++];
    if (ml >= 2 * (a + b)) ml = 0;
  }
  set<char> anw;
  int h = l % (a + b);
  if (h >= a && a > b) {
    set<char> vis;
    for (int i = a + b - h; i < min(int(s2.size()), 2 * a + b - h); i++)
      vis.insert(s2[i]);
    char nc = 'a';
    for (int i = 0; i < 26; i++) {
      if (vis.count('a' + i)) continue;
      nc = 'a' + i;
      break;
    }
    for (int i = 0; i < a + b - h; i++) s2[i] = nc;
  }
  for (char t : s2) anw.insert(t);
  cout << int(anw.size()) << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  l--, r--;
  if (r - l > 100) {
    cout << 2 * a - min(b, a - 1);
    return 0;
  }
  string s = "";
  for (int i = 0; i < a; i++) s += i + 'a';
  for (int i = 0; i < b; i++) s += s[a - 1];
  set<char> vis;
  for (int i = 1; i <= a; i++) vis.insert(s[int(s.size()) - i]);
  for (int i = 0, c = 0; i < 26 && c < a; i++) {
    if (vis.count('a' + i)) continue;
    s += 'a' + i;
    c++;
  }
  for (int i = 0; i < b; i++) s += s.back();
  string s2 = "";
  int ml = l % (2 * (a + b));
  while (r-- >= l) {
    s2 += s[ml++];
    if (ml >= 2 * (a + b)) ml = 0;
  }
  set<char> anw;
  int h = l % (a + b);
  if (h >= a && a > b) {
    set<char> vis;
    for (int i = a + b - h; i < min(int(s2.size()), 2 * a + b - h); i++)
      vis.insert(s2[i]);
    char nc = 'a';
    for (int i = 0; i < 26; i++) {
      if (vis.count('a' + i)) continue;
      nc = 'a' + i;
      break;
    }
    for (int i = 0; i < a + b - h; i++) s2[i] = nc;
  }
  for (char t : s2) anw.insert(t);
  cout << int(anw.size()) << endl;
}

```