#include <bits/stdc++.h>
using namespace std;
int n, a, b, l, r, bo[1000], t1, t2;
string s;
int main() {
  scanf("%d%d%d%d", &a, &b, &l, &r);
  if (l <= a && r <= a) {
    printf("%d", r - l + 1);
    return 0;
  }
  if (l <= a && r <= a + b) {
    printf("%d", a - l + 1);
    return 0;
  }
  if (a <= l && r <= a + b) {
    printf("1");
    return 0;
  }
  if (a <= b) {
    if (r - l > a + b) {
      printf("%d", a + 1);
      return 0;
    }
    if (r - l == a + b) {
      printf("%d", a);
      return 0;
    }
    t1 = l % (a + b);
    t2 = r % (a + b);
    if (1 <= t1 && t1 <= a && 1 <= t2 && t2 <= a) {
      if (r - l < a)
        printf("%d", r - l + 1);
      else
        printf("%d", min(a - t1 + 2, a) + 1);
      return 0;
    }
    if (1 <= t1 && t1 <= a && (a < t2 || t2 == 0)) {
      printf("%d", a - t1 + 1);
      return 0;
    }
    if ((a < t1 || t1 == 0) && 1 <= t2 && t2 <= a) {
      printf("%d", t2 + 1);
      return 0;
    }
    if ((a < t1 || t1 == 0) && (a < t2 || t2 == 0)) {
      if (r - l < b)
        printf("%d", 1);
      else
        printf("%d", a + 1);
    }
  }
  if (a > b) {
    if (r - l >= 2 * (a + b)) {
      printf("%d", a + a - b);
      return 0;
    }
    t1 = l % (2 * (a + b));
    t2 = r % (2 * (a + b));
    if (a < t1 && t1 <= a + b && 1 <= t2 && t2 <= a) {
      printf("%d", a - b + t2);
      return 0;
    }
    for (int i = 1; i <= a; ++i) s += 'a' + i - 1;
    for (int i = 1; i <= b; ++i) s += 'a' + a - 1;
    for (int i = 1; i <= b; ++i) s += 'a' + i - 1;
    for (int i = 1; i <= a - b; ++i) s += 'a' + a - 1 + i;
    for (int i = 1; i <= b; ++i) s += 'a' + a;
    s = s + s;
    s = " " + s + s;
    if (t1 == 0) t1 += 2 * (a + b);
    if (t1 > t2) t2 += 2 * (a + b);
    int ans = 0;
    for (int i = t1; i <= t2; ++i)
      if (!bo[(int)s[i] - 'a']) bo[(int)s[i] - 'a'] = 1, ans++;
    printf("%d", ans);
  }
}

### Prompt

In CPP, your task is to solve 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 n, a, b, l, r, bo[1000], t1, t2;
string s;
int main() {
  scanf("%d%d%d%d", &a, &b, &l, &r);
  if (l <= a && r <= a) {
    printf("%d", r - l + 1);
    return 0;
  }
  if (l <= a && r <= a + b) {
    printf("%d", a - l + 1);
    return 0;
  }
  if (a <= l && r <= a + b) {
    printf("1");
    return 0;
  }
  if (a <= b) {
    if (r - l > a + b) {
      printf("%d", a + 1);
      return 0;
    }
    if (r - l == a + b) {
      printf("%d", a);
      return 0;
    }
    t1 = l % (a + b);
    t2 = r % (a + b);
    if (1 <= t1 && t1 <= a && 1 <= t2 && t2 <= a) {
      if (r - l < a)
        printf("%d", r - l + 1);
      else
        printf("%d", min(a - t1 + 2, a) + 1);
      return 0;
    }
    if (1 <= t1 && t1 <= a && (a < t2 || t2 == 0)) {
      printf("%d", a - t1 + 1);
      return 0;
    }
    if ((a < t1 || t1 == 0) && 1 <= t2 && t2 <= a) {
      printf("%d", t2 + 1);
      return 0;
    }
    if ((a < t1 || t1 == 0) && (a < t2 || t2 == 0)) {
      if (r - l < b)
        printf("%d", 1);
      else
        printf("%d", a + 1);
    }
  }
  if (a > b) {
    if (r - l >= 2 * (a + b)) {
      printf("%d", a + a - b);
      return 0;
    }
    t1 = l % (2 * (a + b));
    t2 = r % (2 * (a + b));
    if (a < t1 && t1 <= a + b && 1 <= t2 && t2 <= a) {
      printf("%d", a - b + t2);
      return 0;
    }
    for (int i = 1; i <= a; ++i) s += 'a' + i - 1;
    for (int i = 1; i <= b; ++i) s += 'a' + a - 1;
    for (int i = 1; i <= b; ++i) s += 'a' + i - 1;
    for (int i = 1; i <= a - b; ++i) s += 'a' + a - 1 + i;
    for (int i = 1; i <= b; ++i) s += 'a' + a;
    s = s + s;
    s = " " + s + s;
    if (t1 == 0) t1 += 2 * (a + b);
    if (t1 > t2) t2 += 2 * (a + b);
    int ans = 0;
    for (int i = t1; i <= t2; ++i)
      if (!bo[(int)s[i] - 'a']) bo[(int)s[i] - 'a'] = 1, ans++;
    printf("%d", ans);
  }
}

```