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;
vector<int> primes;
vector<bool> isPrime;
void primesSieve(int n) {
  isPrime.resize(n + 1, true);
  isPrime[0] = isPrime[1] = false;
  primes.push_back(2);
  for (int i = 4; i <= n; i += 2) isPrime[i] = false;
  int limit = sqrt(n);
  for (int i = 3; i <= n; i += 2) {
    if (isPrime[i]) {
      primes.push_back(i);
      if (i <= limit)
        for (int j = i * i; j <= n; j += 2 * i) isPrime[j] = false;
    }
  }
}
int main() {
  int n = 0, m = 1;
  cin >> n;
  primesSieve(1000000);
  while (true) {
    if (!isPrime[n * m + 1])
      break;
    else
      m++;
  }
  cout << m << endl;
  return 0;
}