#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  int a, b, l, r;
  while (cin >> a >> b >> l >> r) {
    int x = (a + b) * 2;
    char ar[50];
    memset(ar, 0, sizeof(ar));
    for (int i = 0; i < a; i++) ar[i] = 'a' + i;
    for (int i = 0; i < b; i++) ar[i + a] = ar[a - 1];
    int tp;
    if (b >= a)
      tp = 1;
    else
      tp = a - b;
    for (int i = 0; i < a - tp; i++) ar[i + a + b] = 'a' + i;
    for (int i = 0; i < tp; i++) ar[i + 2 * a + b - tp] = i + 'm';
    for (int i = 0; i < b; i++) ar[i + 2 * a + b] = ar[2 * a + b - 1];
    if (r - l >= x) {
      cout << tp + a << endl;
    } else {
      if (r - l + 1 == x - 1 && l % x == a + 1) {
        cout << tp + a - 1 << endl;
        continue;
      }
      l = (l - 1) % x;
      r = (r - 1) % x;
      int i = l;
      int vis[300];
      memset(vis, 0, sizeof(vis));
      int res = 0;
      while (i != r) {
        if (vis[ar[i]] == 0) {
          res++;
          vis[ar[i]] = 1;
        }
        i++;
        if (i >= x) {
          i = 0;
        }
      }
      if (vis[ar[i]] == 0) {
        res++;
        vis[ar[i]] = 1;
      }
      cout << res << endl;
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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 main() {
  ios::sync_with_stdio(false);
  int a, b, l, r;
  while (cin >> a >> b >> l >> r) {
    int x = (a + b) * 2;
    char ar[50];
    memset(ar, 0, sizeof(ar));
    for (int i = 0; i < a; i++) ar[i] = 'a' + i;
    for (int i = 0; i < b; i++) ar[i + a] = ar[a - 1];
    int tp;
    if (b >= a)
      tp = 1;
    else
      tp = a - b;
    for (int i = 0; i < a - tp; i++) ar[i + a + b] = 'a' + i;
    for (int i = 0; i < tp; i++) ar[i + 2 * a + b - tp] = i + 'm';
    for (int i = 0; i < b; i++) ar[i + 2 * a + b] = ar[2 * a + b - 1];
    if (r - l >= x) {
      cout << tp + a << endl;
    } else {
      if (r - l + 1 == x - 1 && l % x == a + 1) {
        cout << tp + a - 1 << endl;
        continue;
      }
      l = (l - 1) % x;
      r = (r - 1) % x;
      int i = l;
      int vis[300];
      memset(vis, 0, sizeof(vis));
      int res = 0;
      while (i != r) {
        if (vis[ar[i]] == 0) {
          res++;
          vis[ar[i]] = 1;
        }
        i++;
        if (i >= x) {
          i = 0;
        }
      }
      if (vis[ar[i]] == 0) {
        res++;
        vis[ar[i]] = 1;
      }
      cout << res << endl;
    }
  }
  return 0;
}

```