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;
bool isPrime(int num);
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= 1000; i++) {
    if (!(isPrime(n * i + 1))) {
      cout << i;
      break;
    }
  }
  return 0;
}
bool isPrime(int num) {
  if (num <= 1)
    return false;
  else if (num == 2)
    return true;
  else if (num % 2 == 0)
    return false;
  else {
    bool prime = true;
    int divisor = 3;
    double num_d = static_cast<double>(num);
    int upperLimit = static_cast<int>(sqrt(num_d) + 1);
    while (divisor <= upperLimit) {
      if (num % divisor == 0) prime = false;
      divisor += 2;
    }
    return prime;
  }
}