Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int sieve[1000005];
void prime() {
  sieve[1] = 0;
  for (int i = 2; i * i < 1000005; i++) {
    if (sieve[i] == 1) {
      for (int j = i * i; j < 1000005; j += i) sieve[j] = 0;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    int n;
    cin >> n;
    for (int i = 0; i < 1000005; i++) sieve[i] = 1;
    prime();
    if (n % 2 == 1) {
      cout << 3;
    } else {
      for (int i = 1; i <= 1000; i++) {
        int p = n * i + 1;
        if (!sieve[p]) {
          cout << i;
          exit(0);
        }
      }
    }
  }
  return 0;
}