#include <bits/stdc++.h>
using namespace std;
int a, b, l, r;
bool life[30];
int num[60];
int go(int x1, int x2) {
  int now, x;
  memset(life, 0, sizeof(life));
  for (int i = 1; i <= a; i++) num[i] = i;
  for (int i = 1; i <= b; i++) {
    num[i + a] = x1;
    num[i + a * 2 + b] = x2;
  }
  for (int i = b + 1; i <= a + b; i++) life[num[i]] = 1;
  now = 1;
  for (int i = a + b + 1; i <= a * 2 + b; i++) {
    while (life[now]) now++;
    num[i] = now;
    now++;
  }
  now = 0;
  memset(life, 0, sizeof(life));
  for (int i = l; i <= r; i++) {
    x = i % (2 * a + 2 * b);
    if (x == 0) x = 2 * a + 2 * b;
    if (life[num[x]] == 0) {
      now++;
      life[num[x]] = 1;
    }
  }
  return now;
}
int main() {
  int ans;
  scanf("%d%d%d%d", &a, &b, &l, &r);
  if (r - l + 1 >= 2 * a + 2 * b) {
    if (b >= a)
      ans = a + 1;
    else
      ans = 2 * a - b;
  } else {
    l = l % (2 * a + 2 * b);
    if (l == 0) l = 2 * a + 2 * b;
    r = r % (2 * a + 2 * b);
    if (r < l) r += 2 * a + 2 * b;
    ans = 26;
    for (int i = 1; i <= 26; i++)
      for (int j = a + 1; j <= 26; j++) ans = min(ans, go(i, j));
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate a cpp solution to 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 a, b, l, r;
bool life[30];
int num[60];
int go(int x1, int x2) {
  int now, x;
  memset(life, 0, sizeof(life));
  for (int i = 1; i <= a; i++) num[i] = i;
  for (int i = 1; i <= b; i++) {
    num[i + a] = x1;
    num[i + a * 2 + b] = x2;
  }
  for (int i = b + 1; i <= a + b; i++) life[num[i]] = 1;
  now = 1;
  for (int i = a + b + 1; i <= a * 2 + b; i++) {
    while (life[now]) now++;
    num[i] = now;
    now++;
  }
  now = 0;
  memset(life, 0, sizeof(life));
  for (int i = l; i <= r; i++) {
    x = i % (2 * a + 2 * b);
    if (x == 0) x = 2 * a + 2 * b;
    if (life[num[x]] == 0) {
      now++;
      life[num[x]] = 1;
    }
  }
  return now;
}
int main() {
  int ans;
  scanf("%d%d%d%d", &a, &b, &l, &r);
  if (r - l + 1 >= 2 * a + 2 * b) {
    if (b >= a)
      ans = a + 1;
    else
      ans = 2 * a - b;
  } else {
    l = l % (2 * a + 2 * b);
    if (l == 0) l = 2 * a + 2 * b;
    r = r % (2 * a + 2 * b);
    if (r < l) r += 2 * a + 2 * b;
    ans = 26;
    for (int i = 1; i <= 26; i++)
      for (int j = a + 1; j <= 26; j++) ans = min(ans, go(i, j));
  }
  printf("%d\n", ans);
  return 0;
}

```