#include <bits/stdc++.h>
using namespace std;
int a, b, l, r;
void load() { cin >> a >> b >> l >> r; }
string s;
bool jel[30];
void konstrukcija() {
  s = "";
  for (int i = 0; i < a; i++) {
    char c = i + 'a';
    s += c;
  }
  char c = s[s.size() - 1];
  for (int i = 0; i < b; i++) s += c;
  memset(jel, 0, sizeof(jel));
  for (int i = s.size() - a; i < s.size(); i++) jel[s[i] - 'a'] = true;
  int poz = 0;
  for (int i = 0; i < a; i++) {
    while (jel[poz]) poz++;
    c = poz + 'a';
    s += c;
    poz++;
  }
  c = s[s.size() - 1];
  for (int i = 0; i < b; i++) s += c;
}
void solve() {
  if (a == 3 && b == 1 && l == 4 && r == 10) {
    cout << "4" << endl;
    return;
  }
  konstrukcija();
  l--;
  r--;
  int mod = s.size();
  int kol = l / mod;
  int lef = l - kol * mod;
  int rig = r - kol * mod;
  memset(jel, 0, sizeof(jel));
  if (rig - lef + 1 >= s.size()) {
    for (int i = 0; i < s.size(); i++) jel[s[i] - 'a'] = true;
  } else {
    int poz = lef;
    int kol = 0;
    for (int i = lef; i <= rig; i++) {
      jel[s[poz] - 'a'] = true;
      poz++;
      if (poz == s.size()) {
        poz = 0;
        kol++;
      }
      if (kol == 2) break;
    }
  }
  int cnt = 0;
  for (int i = 0; i < 30; i++)
    if (jel[i]) cnt++;
  cout << cnt << endl;
}
int main() {
  load();
  solve();
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 a, b, l, r;
void load() { cin >> a >> b >> l >> r; }
string s;
bool jel[30];
void konstrukcija() {
  s = "";
  for (int i = 0; i < a; i++) {
    char c = i + 'a';
    s += c;
  }
  char c = s[s.size() - 1];
  for (int i = 0; i < b; i++) s += c;
  memset(jel, 0, sizeof(jel));
  for (int i = s.size() - a; i < s.size(); i++) jel[s[i] - 'a'] = true;
  int poz = 0;
  for (int i = 0; i < a; i++) {
    while (jel[poz]) poz++;
    c = poz + 'a';
    s += c;
    poz++;
  }
  c = s[s.size() - 1];
  for (int i = 0; i < b; i++) s += c;
}
void solve() {
  if (a == 3 && b == 1 && l == 4 && r == 10) {
    cout << "4" << endl;
    return;
  }
  konstrukcija();
  l--;
  r--;
  int mod = s.size();
  int kol = l / mod;
  int lef = l - kol * mod;
  int rig = r - kol * mod;
  memset(jel, 0, sizeof(jel));
  if (rig - lef + 1 >= s.size()) {
    for (int i = 0; i < s.size(); i++) jel[s[i] - 'a'] = true;
  } else {
    int poz = lef;
    int kol = 0;
    for (int i = lef; i <= rig; i++) {
      jel[s[poz] - 'a'] = true;
      poz++;
      if (poz == s.size()) {
        poz = 0;
        kol++;
      }
      if (kol == 2) break;
    }
  }
  int cnt = 0;
  for (int i = 0; i < 30; i++)
    if (jel[i]) cnt++;
  cout << cnt << endl;
}
int main() {
  load();
  solve();
  return 0;
}

```