#include <bits/stdc++.h>
using namespace std;
const int INF = (int)1e9;
void solve() {
  int alph[26], alph1[26];
  for (int i = 0; i < 26; i++) alph[i] = alph1[i] = 0;
  int a, b, l, r, dif, ans1 = INF;
  string s, s1;
  cin >> a >> b >> l >> r;
  --l;
  --r;
  char d;
  for (char c = 'a'; c - 'a' < a; c++) {
    if (a > b && c - 'a' == b) d = c;
    s += c;
    if (a > b) s1 += c;
    if (c - 'a' == a - 1) {
      for (int i = 0; i < b; i++) {
        s += c;
        if (a > b) s1 += d;
      }
    }
  }
  int sz = s.length(), times = 0, times1 = 0;
  for (int i = sz - 1; times < a; i--) {
    alph[s[i] - 'a']++;
    if (a > b) alph1[s1[i] - 'a']++;
    times++;
  }
  times = 0;
  if (!(a > b)) times1 = INF;
  char c = 'a';
  for (int i = 0; times < a || times1 < a; i++) {
    if (times1 == b) d = c;
    if (alph[i] == 0) {
      s += c;
      times++;
      if (times == a) {
        for (int j = 0; j < b; j++) {
          s += c;
        }
      }
    }
    if (alph1[i] == 0) {
      if (a > b) s1 += c;
      times1++;
      if (times1 == a) {
        for (int j = 0; j < b; j++) {
          s1 += d;
        }
      }
    }
    c++;
  }
  for (int i = 0; i < 26; i++) alph[i] = alph1[i] = 0;
  dif = l;
  l = l % (2 * (a + b));
  dif -= l;
  r -= dif;
  int idx = l;
  for (int i = 0; i < min(2 * (a + b), r - l + 1); i++) {
    alph[s[idx % (2 * (a + b))] - 'a']++;
    if (a > b) alph1[s1[idx % (2 * (a + b))] - 'a']++;
    idx++;
  }
  int answer = 0;
  if (a > b) ans1 = 0;
  for (int i = 0; i < 26; i++) {
    if (alph[i]) answer++;
    if (alph1[i]) ans1++;
  }
  cout << min(answer, ans1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
const int INF = (int)1e9;
void solve() {
  int alph[26], alph1[26];
  for (int i = 0; i < 26; i++) alph[i] = alph1[i] = 0;
  int a, b, l, r, dif, ans1 = INF;
  string s, s1;
  cin >> a >> b >> l >> r;
  --l;
  --r;
  char d;
  for (char c = 'a'; c - 'a' < a; c++) {
    if (a > b && c - 'a' == b) d = c;
    s += c;
    if (a > b) s1 += c;
    if (c - 'a' == a - 1) {
      for (int i = 0; i < b; i++) {
        s += c;
        if (a > b) s1 += d;
      }
    }
  }
  int sz = s.length(), times = 0, times1 = 0;
  for (int i = sz - 1; times < a; i--) {
    alph[s[i] - 'a']++;
    if (a > b) alph1[s1[i] - 'a']++;
    times++;
  }
  times = 0;
  if (!(a > b)) times1 = INF;
  char c = 'a';
  for (int i = 0; times < a || times1 < a; i++) {
    if (times1 == b) d = c;
    if (alph[i] == 0) {
      s += c;
      times++;
      if (times == a) {
        for (int j = 0; j < b; j++) {
          s += c;
        }
      }
    }
    if (alph1[i] == 0) {
      if (a > b) s1 += c;
      times1++;
      if (times1 == a) {
        for (int j = 0; j < b; j++) {
          s1 += d;
        }
      }
    }
    c++;
  }
  for (int i = 0; i < 26; i++) alph[i] = alph1[i] = 0;
  dif = l;
  l = l % (2 * (a + b));
  dif -= l;
  r -= dif;
  int idx = l;
  for (int i = 0; i < min(2 * (a + b), r - l + 1); i++) {
    alph[s[idx % (2 * (a + b))] - 'a']++;
    if (a > b) alph1[s1[idx % (2 * (a + b))] - 'a']++;
    idx++;
  }
  int answer = 0;
  if (a > b) ans1 = 0;
  for (int i = 0; i < 26; i++) {
    if (alph[i]) answer++;
    if (alph1[i]) ans1++;
  }
  cout << min(answer, ans1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```