#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long expo(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y % 2) res = (res * x % mod) % mod;
    x = (x * x) % mod;
    y /= 2;
  }
  return res;
}
long long inv(long long x) { return expo(x, mod - 2); }
long long add(long long x, long long y) { return (x + y) % mod; }
long long sub(long long x, long long y) { return (x - y) % mod; }
long long mul(long long x, long long y) {
  return ((x % mod) * (y % mod)) % mod;
}
long long n = 100001;
vector<bool> is_prime(n, true);
void sei() {
  is_prime[1] = false;
  is_prime[0] = false;
  for (long long i = 2; i * i <= n; i++) {
    if (is_prime[i]) {
      for (long long j = i * i; j <= n; j += i) is_prime[j] = false;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  sei();
  long long k;
  cin >> k;
  if (k == 1) {
    cout << "3\n";
    return 0;
  }
  if (k % 2)
    cout << "1\n";
  else {
    long long m = 1;
    while (1) {
      long long temp = k * m + 1;
      if (!is_prime[temp]) {
        cout << m << "\n";
        break;
      }
      m++;
    }
  }
}

### Prompt

Please formulate a cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long expo(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y % 2) res = (res * x % mod) % mod;
    x = (x * x) % mod;
    y /= 2;
  }
  return res;
}
long long inv(long long x) { return expo(x, mod - 2); }
long long add(long long x, long long y) { return (x + y) % mod; }
long long sub(long long x, long long y) { return (x - y) % mod; }
long long mul(long long x, long long y) {
  return ((x % mod) * (y % mod)) % mod;
}
long long n = 100001;
vector<bool> is_prime(n, true);
void sei() {
  is_prime[1] = false;
  is_prime[0] = false;
  for (long long i = 2; i * i <= n; i++) {
    if (is_prime[i]) {
      for (long long j = i * i; j <= n; j += i) is_prime[j] = false;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  sei();
  long long k;
  cin >> k;
  if (k == 1) {
    cout << "3\n";
    return 0;
  }
  if (k % 2)
    cout << "1\n";
  else {
    long long m = 1;
    while (1) {
      long long temp = k * m + 1;
      if (!is_prime[temp]) {
        cout << m << "\n";
        break;
      }
      m++;
    }
  }
}

```