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 prime[3000000], nPrime;
int mark[10000002];
void sieve(int n) {
  int i, j, limit = sqrt(n * 1.) + 2;
  mark[1] = 1;
  for (int i = 4; i <= n; i += 2) mark[i] = 1;
  prime[nPrime++] = 2;
  for (i = 3; i <= n; i++) {
    if (!mark[i]) {
      prime[nPrime++] = i;
      if (i <= limit) {
        for (j = i * i; j <= n; j += i * 2) {
          mark[j] = 1;
        }
      }
    }
  }
}
int main() {
  int n;
  sieve(10000);
  while (cin >> n) {
    for (int i = 1; i <= 1001; i++) {
      int x = n * i;
      if (mark[x + 1] == 1) {
        cout << i << endl;
        return 0;
      }
    }
  }
}