#include <bits/stdc++.h>
using namespace std;
const double sn = 1e-6;
int a, b, l, r;
int main() {
  scanf("%d%d%d%d", &a, &b, &l, &r);
  if (r - l + 1 >= (a + b) * 2) {
    if (b >= a) {
      printf("%d\n", a + 1);
    } else {
      printf("%d\n", a + (a - b));
    }
  } else {
    int t = 1, v2 = r % ((a + b) * 2), co = 0, temp = 1;
    vector<int> ve = {};
    bool last = false;
    int t2 = 1, t3 = 1;
    while (true) {
      if (t2 == v2 && t3 >= (r - l) + 1) break;
      t2 = (t2 + 1) % (2 * (a + b));
      t3++;
    }
    while (true) {
      if ((t >= 1 && t <= a) || (t >= a + b + 1 && t <= 2 * a + b)) {
        last = false;
        int dig = 1;
        while (true) {
          bool val = true;
          for (int i = ve.size() - 1; i >= max((int)ve.size() - a - co, 0);
               i--) {
            if (ve[i] == dig) {
              val = false;
              break;
            }
          }
          if (val)
            break;
          else
            dig++;
        }
        ve.push_back(dig);
        co++;
      } else {
        co = 0;
        if (b >= a || last) {
          ve.push_back(ve[ve.size() - 1]);
        } else {
          int sz = ve.size();
          int tv = t3 - (r - l + 1);
          if (sz - (a - b) >= tv || sz <= tv) {
            ve.push_back(ve[sz - (a - b)]);
          } else {
            ve.push_back(ve[tv]);
          }
        }
        last = true;
      }
      if (t == v2 && temp >= (r - l) + 1) break;
      t = (t + 1) % ((a + b) * 2);
      temp++;
    }
    bool vis[30] = {};
    int ans = 0, sz = ve.size();
    for (int i = sz - 1; i >= sz - (r - l + 1); i--) {
      if (!vis[ve[i]]) {
        vis[ve[i]] = true;
        ans++;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const double sn = 1e-6;
int a, b, l, r;
int main() {
  scanf("%d%d%d%d", &a, &b, &l, &r);
  if (r - l + 1 >= (a + b) * 2) {
    if (b >= a) {
      printf("%d\n", a + 1);
    } else {
      printf("%d\n", a + (a - b));
    }
  } else {
    int t = 1, v2 = r % ((a + b) * 2), co = 0, temp = 1;
    vector<int> ve = {};
    bool last = false;
    int t2 = 1, t3 = 1;
    while (true) {
      if (t2 == v2 && t3 >= (r - l) + 1) break;
      t2 = (t2 + 1) % (2 * (a + b));
      t3++;
    }
    while (true) {
      if ((t >= 1 && t <= a) || (t >= a + b + 1 && t <= 2 * a + b)) {
        last = false;
        int dig = 1;
        while (true) {
          bool val = true;
          for (int i = ve.size() - 1; i >= max((int)ve.size() - a - co, 0);
               i--) {
            if (ve[i] == dig) {
              val = false;
              break;
            }
          }
          if (val)
            break;
          else
            dig++;
        }
        ve.push_back(dig);
        co++;
      } else {
        co = 0;
        if (b >= a || last) {
          ve.push_back(ve[ve.size() - 1]);
        } else {
          int sz = ve.size();
          int tv = t3 - (r - l + 1);
          if (sz - (a - b) >= tv || sz <= tv) {
            ve.push_back(ve[sz - (a - b)]);
          } else {
            ve.push_back(ve[tv]);
          }
        }
        last = true;
      }
      if (t == v2 && temp >= (r - l) + 1) break;
      t = (t + 1) % ((a + b) * 2);
      temp++;
    }
    bool vis[30] = {};
    int ans = 0, sz = ve.size();
    for (int i = sz - 1; i >= sz - (r - l + 1); i--) {
      if (!vis[ve[i]]) {
        vis[ve[i]] = true;
        ans++;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

```