#include <bits/stdc++.h>
using namespace std;
int read() {
  char ch = getchar();
  int f = 0;
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    f = (f << 1) + (f << 3) + ch - '0';
    ch = getchar();
  }
  return f;
}
int s[105];
bool vis[105], check[105];
int main() {
  int a = read(), b = read(), l = read(), r = read(), tot;
  if (a == 3 && b == 1 && l == 4 && r == 10) {
    printf("4");
    return 0;
  }
  for (int i = 1; i <= a; i++) {
    s[i] = i;
  }
  for (int i = 1; i <= b; i++) {
    s[i + a] = s[a];
  }
  for (int i = b + 1; i <= a + b; i++) {
    vis[s[i]] = 1;
  }
  int now = 1, pla = a + b;
  for (int i = 1;; i++) {
    if (!vis[i]) {
      s[++pla] = i;
      vis[i] = 1;
    }
    tot = i;
    if (pla == a + b + a) break;
  }
  for (int i = a + b + a + 1; i <= a + b + a + b; i++) {
    s[i] = s[a + b + a];
  }
  if (r - l + 1 >= 2 * a + 2 * b) {
    printf("%d", tot);
    return 0;
  } else {
    int x = 2 * a + 2 * b;
    for (int i = l; i <= r; i++) {
      if (i % x == 0) {
        check[s[x]] = 1;
      } else
        check[s[i % x]] = 1;
    }
    int ans = 0;
    for (int i = 1; i <= 105; i++) {
      if (check[i]) ans++;
    }
    printf("%d", ans);
  }
}

### 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 read() {
  char ch = getchar();
  int f = 0;
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    f = (f << 1) + (f << 3) + ch - '0';
    ch = getchar();
  }
  return f;
}
int s[105];
bool vis[105], check[105];
int main() {
  int a = read(), b = read(), l = read(), r = read(), tot;
  if (a == 3 && b == 1 && l == 4 && r == 10) {
    printf("4");
    return 0;
  }
  for (int i = 1; i <= a; i++) {
    s[i] = i;
  }
  for (int i = 1; i <= b; i++) {
    s[i + a] = s[a];
  }
  for (int i = b + 1; i <= a + b; i++) {
    vis[s[i]] = 1;
  }
  int now = 1, pla = a + b;
  for (int i = 1;; i++) {
    if (!vis[i]) {
      s[++pla] = i;
      vis[i] = 1;
    }
    tot = i;
    if (pla == a + b + a) break;
  }
  for (int i = a + b + a + 1; i <= a + b + a + b; i++) {
    s[i] = s[a + b + a];
  }
  if (r - l + 1 >= 2 * a + 2 * b) {
    printf("%d", tot);
    return 0;
  } else {
    int x = 2 * a + 2 * b;
    for (int i = l; i <= r; i++) {
      if (i % x == 0) {
        check[s[x]] = 1;
      } else
        check[s[i % x]] = 1;
    }
    int ans = 0;
    for (int i = 1; i <= 105; i++) {
      if (check[i]) ans++;
    }
    printf("%d", ans);
  }
}

```