#include <bits/stdc++.h>
using namespace std;
int main(int argc, char *argv[]) {
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  {
    int period = a + b;
    int res;
    int start = l / period * period - 10 * period;
    if (start < 0) start = 0;
    l--;
    r--;
    l -= start;
    r -= start;
    if (r >= l + 500) {
      r = l + 500;
    }
    char s[1000];
    int lastSend[26], lastTime = 7;
    for (int i = 0; i < 26; i++) lastSend[i] = 0;
    for (int i = 0; i + period < 1000; i += period) {
      for (int j = 0, c = 0; j < a; j++) {
        while (lastSend[c] == lastTime) c++;
        s[i + j] = c++;
      }
      for (int j = a; j < period; j++) {
        s[i + j] = s[min(i + a - 1, i + period - a)];
      }
      lastTime++;
      for (int j = period - a; j < period; j++) {
        lastSend[s[i + j]] = lastTime;
      }
    }
    lastTime++;
    res = 0;
    for (int i = l; i <= r; i++) {
      if (lastSend[s[i]] != lastTime) {
        res++;
        lastSend[s[i]] = lastTime;
      }
    }
    lastTime++;
    for (int i = 0; i + period < 1000; i += period) {
      for (int j = 0, c = 0; j < a; j++) {
        while (lastSend[c] == lastTime) c++;
        s[i + j] = c++;
      }
      for (int j = a; j < period; j++) {
        s[i + j] = s[i + a - 1];
      }
      lastTime++;
      for (int j = period - a; j < period; j++) {
        lastSend[s[i + j]] = lastTime;
      }
    }
    lastTime++;
    int res2 = 0;
    for (int i = l; i <= r; i++) {
      if (lastSend[s[i]] != lastTime) {
        res2++;
        lastSend[s[i]] = lastTime;
      }
    }
    cout << min(res, res2) << endl;
  }
}

### 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 main(int argc, char *argv[]) {
  int a, b, l, r;
  cin >> a >> b >> l >> r;
  {
    int period = a + b;
    int res;
    int start = l / period * period - 10 * period;
    if (start < 0) start = 0;
    l--;
    r--;
    l -= start;
    r -= start;
    if (r >= l + 500) {
      r = l + 500;
    }
    char s[1000];
    int lastSend[26], lastTime = 7;
    for (int i = 0; i < 26; i++) lastSend[i] = 0;
    for (int i = 0; i + period < 1000; i += period) {
      for (int j = 0, c = 0; j < a; j++) {
        while (lastSend[c] == lastTime) c++;
        s[i + j] = c++;
      }
      for (int j = a; j < period; j++) {
        s[i + j] = s[min(i + a - 1, i + period - a)];
      }
      lastTime++;
      for (int j = period - a; j < period; j++) {
        lastSend[s[i + j]] = lastTime;
      }
    }
    lastTime++;
    res = 0;
    for (int i = l; i <= r; i++) {
      if (lastSend[s[i]] != lastTime) {
        res++;
        lastSend[s[i]] = lastTime;
      }
    }
    lastTime++;
    for (int i = 0; i + period < 1000; i += period) {
      for (int j = 0, c = 0; j < a; j++) {
        while (lastSend[c] == lastTime) c++;
        s[i + j] = c++;
      }
      for (int j = a; j < period; j++) {
        s[i + j] = s[i + a - 1];
      }
      lastTime++;
      for (int j = period - a; j < period; j++) {
        lastSend[s[i + j]] = lastTime;
      }
    }
    lastTime++;
    int res2 = 0;
    for (int i = l; i <= r; i++) {
      if (lastSend[s[i]] != lastTime) {
        res2++;
        lastSend[s[i]] = lastTime;
      }
    }
    cout << min(res, res2) << endl;
  }
}

```