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;
void k_007() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long prime(long long n) {
  long long flag = 1;
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      flag = 0;
      break;
    }
  }
  return flag;
}
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return ((power(x, y / 2) % 1000000007) * (power(x, y / 2) % 1000000007)) %
           1000000007;
  else
    return ((x % 1000000007) * (power(x, y / 2) % 1000000007) *
            (power(x, y / 2) % 1000000007)) %
           1000000007;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int32_t main() {
  k_007();
  long long n;
  cin >> n;
  if (n & 1) {
    if (n == 1) {
      cout << 3 << endl;
    } else {
      cout << 1 << endl;
    }
  } else {
    long long m = 1;
    while (prime(n * m + 1)) {
      m++;
    }
    cout << m << endl;
  }
}