#include <bits/stdc++.h>
using namespace std;
const int N = 45000;
vector<int> p;
bool isPrime(int n) {
  for (int i = 0; p[i] < n && i < p.size(); i++) {
    if (n % p[i] == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  bool pp[N];
  memset(pp, 1, N * sizeof(bool));
  pp[0] = pp[1] = false;
  for (long long i = 2; i < N; i++) {
    if (pp[i]) {
      p.push_back(i);
      for (long long j = i * 2; j < N; j += i) {
        pp[j] = false;
      }
    }
  }
  int n;
  while (cin >> n) {
    int k = 0;
    if (n == 2 || isPrime(n)) {
      k = 1;
    } else if (n % 2 == 0 || isPrime(n - 2)) {
      k = 2;
    } else {
      k = 3;
    }
    cout << k << endl;
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.

As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.

Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.

Output

Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.

Examples

Input

4


Output

2


Input

27


Output

3

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 45000;
vector<int> p;
bool isPrime(int n) {
  for (int i = 0; p[i] < n && i < p.size(); i++) {
    if (n % p[i] == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  bool pp[N];
  memset(pp, 1, N * sizeof(bool));
  pp[0] = pp[1] = false;
  for (long long i = 2; i < N; i++) {
    if (pp[i]) {
      p.push_back(i);
      for (long long j = i * 2; j < N; j += i) {
        pp[j] = false;
      }
    }
  }
  int n;
  while (cin >> n) {
    int k = 0;
    if (n == 2 || isPrime(n)) {
      k = 1;
    } else if (n % 2 == 0 || isPrime(n - 2)) {
      k = 2;
    } else {
      k = 3;
    }
    cout << k << endl;
  }
  return 0;
}

```