#include <bits/stdc++.h>
using namespace std;
bool check[1000001];
vector<int> prime;
void sieve() {
  check[1] = true;
  for (int i = 2; i <= 1000001; i++) {
    if (check[i] == false) {
      for (int j = 2 * i; j <= 1000001; j += i) {
        check[j] = true;
      }
    }
  }
}
long long int power(long long int n) {
  long long int a = 2;
  long long int res = 1;
  while (n) {
    if (n & 1) {
      res = (res * a);
    }
    n >>= 1;
    a = a * a;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  sieve();
  for (int i = 1; i <= 1000; i++) {
    if (check[n * i + 1]) {
      cout << i << endl;
      return 0;
    }
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer n that for each positive integer m number n·m + 1 is a prime number".

Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any n.

Input

The only number in the input is n (1 ≤ n ≤ 1000) — number from the PolandBall's hypothesis. 

Output

Output such m that n·m + 1 is not a prime number. Your answer will be considered correct if you output any suitable m such that 1 ≤ m ≤ 103. It is guaranteed the the answer exists.

Examples

Input

3


Output

1

Input

4


Output

2

Note

A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.

For the first sample testcase, 3·1 + 1 = 4. We can output 1.

In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, m = 2 is okay since 4·2 + 1 = 9, which is not a prime number.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
bool check[1000001];
vector<int> prime;
void sieve() {
  check[1] = true;
  for (int i = 2; i <= 1000001; i++) {
    if (check[i] == false) {
      for (int j = 2 * i; j <= 1000001; j += i) {
        check[j] = true;
      }
    }
  }
}
long long int power(long long int n) {
  long long int a = 2;
  long long int res = 1;
  while (n) {
    if (n & 1) {
      res = (res * a);
    }
    n >>= 1;
    a = a * a;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  sieve();
  for (int i = 1; i <= 1000; i++) {
    if (check[n * i + 1]) {
      cout << i << endl;
      return 0;
    }
  }
}

```