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 maxim = 1000001;
vector<int> primes;
void seive() {
  vector<bool> isprime(maxim + 1, true);
  for (int i = 2; i * i <= maxim; i++) {
    if (isprime[i]) {
      for (int j = 2 * i; j <= maxim; j += i) {
        isprime[j] = false;
      }
    }
  }
  for (int i = 2; i <= maxim; i++) {
    if (isprime[i]) {
      primes.push_back(i);
    }
  }
}
int main() {
  int n, ans = 0;
  cin >> n;
  seive();
  for (int i = 1; i <= 1000; i++) {
    int z = n * i + 1;
    if (!binary_search(primes.begin(), primes.end(), z)) {
      ans = i;
      break;
    }
  }
  cout << ans;
  return 0;
}