#include <bits/stdc++.h>
using namespace std;
int arr[97];
void fill(int l, int r, int v) {
  for (int i = l; i <= r; i++) arr[i] = v;
}
void fill(int l, int r) {
  for (int i = l; i <= r; i++) arr[i] = i;
}
void add(int l, int r) {
  set<int> s;
  for (int i = 1; i <= 26; i++) s.insert(i);
  int a = r - l + 1;
  for (int i = l - a; i <= r - a; i++) s.erase(arr[i]);
  for (int i = l; i <= r; i++) arr[i] = *s.begin(), s.erase(s.begin());
}
int prov(int l, int r) {
  set<int> s;
  for (int i = l; i <= r; i++) s.insert(arr[i]);
  return s.size();
}
int main() {
  int a, b, l, r;
  scanf("%d%d%d%d", &a, &b, &l, &r);
  int d = 2 * (a + b);
  if (r - l + 1 >= d) return printf("%d", max(a + 1, 2 * a - b)), 0;
  r -= (l - 1) / d * d;
  l -= (l - 1) / d * d;
  int m = 26;
  fill(1, a);
  for (int i = 1; i <= 26; i++)
    for (int j = 1; j <= 26; j++)
      for (int k = 1; k <= 26; k++) {
        fill(a + 1, a + b, i);
        add(a + b + 1, 2 * a + b);
        fill(2 * a + b + 1, 2 * a + 2 * b, j);
        add(2 * a + 2 * b + 1, 3 * a + 2 * b);
        fill(3 * a + 2 * b + 1, 3 * a + 3 * b, k);
        add(3 * a + 3 * b + 1, 4 * a + 3 * b);
        fill(4 * a + 3 * b + 1, 4 * a + 4 * b, arr[4 * a + 3 * b]);
        int p = prov(l, r);
        m = min(m, p);
      }
  printf("%d", m);
}

### Prompt

Create a solution in CPP for the following problem:

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 arr[97];
void fill(int l, int r, int v) {
  for (int i = l; i <= r; i++) arr[i] = v;
}
void fill(int l, int r) {
  for (int i = l; i <= r; i++) arr[i] = i;
}
void add(int l, int r) {
  set<int> s;
  for (int i = 1; i <= 26; i++) s.insert(i);
  int a = r - l + 1;
  for (int i = l - a; i <= r - a; i++) s.erase(arr[i]);
  for (int i = l; i <= r; i++) arr[i] = *s.begin(), s.erase(s.begin());
}
int prov(int l, int r) {
  set<int> s;
  for (int i = l; i <= r; i++) s.insert(arr[i]);
  return s.size();
}
int main() {
  int a, b, l, r;
  scanf("%d%d%d%d", &a, &b, &l, &r);
  int d = 2 * (a + b);
  if (r - l + 1 >= d) return printf("%d", max(a + 1, 2 * a - b)), 0;
  r -= (l - 1) / d * d;
  l -= (l - 1) / d * d;
  int m = 26;
  fill(1, a);
  for (int i = 1; i <= 26; i++)
    for (int j = 1; j <= 26; j++)
      for (int k = 1; k <= 26; k++) {
        fill(a + 1, a + b, i);
        add(a + b + 1, 2 * a + b);
        fill(2 * a + b + 1, 2 * a + 2 * b, j);
        add(2 * a + 2 * b + 1, 3 * a + 2 * b);
        fill(3 * a + 2 * b + 1, 3 * a + 3 * b, k);
        add(3 * a + 3 * b + 1, 4 * a + 3 * b);
        fill(4 * a + 3 * b + 1, 4 * a + 4 * b, arr[4 * a + 3 * b]);
        int p = prov(l, r);
        m = min(m, p);
      }
  printf("%d", m);
}

```