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>
inline int in(int &pleasedontbemad) { return scanf("%d", &pleasedontbemad); }
inline int inl(long long &glenda) { return scanf("%lld", &glenda); }
inline int out(int rania) { return printf("%d\n", rania); }
inline int outl(long long yolanda) { return printf("%lld\n", yolanda); }
using namespace std;
inline void base() { ios_base::sync_with_stdio(false); }
inline void tie() { cin.tie(0); }
int m, tmp;
bool vis[1000005];
void sieve() {
  for (long long i = 2; i <= 1000000; i++) {
    if (!vis[i]) {
      for (long long j = (i * i); j <= 1000000; j = j + i) {
        vis[j] = 1;
      }
    }
  }
}
int main() {
  base();
  sieve();
  cin >> m;
  for (int i = 1; i <= 1000; i++) {
    tmp = m * i + 1;
    if (vis[tmp]) {
      cout << i << endl;
      break;
    }
  }
  return 0;
}