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(long long n) {
  if (n == 2 or n == 3)
    return true;
  else if ((n - 1) % 6 == 0 or (n + 1) % 6 == 0)
    return true;
  else
    return false;
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i < 10000; i++) {
    if (isPrime((n * i) + 1))
      continue;
    else {
      cout << i << "\n";
      break;
    }
  }
}
void newSolution() {
  long long n;
  cin >> n;
  if (n == 1)
    cout << 3 << "\n";
  else if (n == 2)
    cout << 4 << "\n";
  else {
    cout << n - 2 << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long testcases = 1;
  while (testcases--) {
    newSolution();
  }
  return 0;
}