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 const INF = 100 + (int)1e9;
long long const INFL = 100 + (long long)1e18;
long double const PI = 3.141592653589793238462643L;
mt19937 tw(960172);
bool is_prime(long long x) {
  for (long long y = 2; y * y <= x; ++y)
    if (x % y == 0) return 0;
  return x > 1;
}
long long rnd(long long x, long long y) {
  static uniform_int_distribution<long long> d;
  return d(tw) % (y - x + 1) + x;
}
long long sqr(int a) { return (long long)a * a; }
template <class T>
T sqr(T const& a) {
  return a * a;
}
long long gcd(long long a, long long b) {
  while (b > 0) {
    long long t = a % b;
    a = b;
    b = t;
  }
  return a;
}
void solve() {
  int n;
  cin >> n;
  int m = 1;
  while (is_prime(n * m + 1)) ++m;
  cout << m << '\n';
}
int main() {
  cout << setprecision(15) << fixed;
  int tcnt = 1;
  for (int test = 1; test <= tcnt; ++test) solve();
}