#include <bits/stdc++.h>
using namespace std;
long long add(long long a, long long b, long long m) {
  long long x = a + b;
  while (x >= m) x -= m;
  return x;
}
long long sub(long long a, long long b, long long m) {
  long long x = a - b;
  while (x < 0) x += m;
  return x;
}
long long mul(long long a, long long b, long long m) {
  long long x = a * b;
  x %= m;
  return x;
}
long long powermod(long long _a, long long _b, long long _m = 1000000007) {
  long long _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a) % _m;
    _b /= 2;
    _a = (_a * _a) % _m;
  }
  return _r;
}
bool prime[1000005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  for (long long i = 0; i < 1000005; i++) prime[i] = true;
  for (long long i = 2; i * i <= 1000005; i++) {
    if (prime[i] == true)
      for (long long j = 2 * i; j <= 1000005; j += i) prime[j] = false;
  }
  long long n;
  cin >> n;
  for (long long i = 1; i < 1001; i++) {
    if (prime[n * i + 1] == false) {
      cout << i << endl;
      return 0;
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
long long add(long long a, long long b, long long m) {
  long long x = a + b;
  while (x >= m) x -= m;
  return x;
}
long long sub(long long a, long long b, long long m) {
  long long x = a - b;
  while (x < 0) x += m;
  return x;
}
long long mul(long long a, long long b, long long m) {
  long long x = a * b;
  x %= m;
  return x;
}
long long powermod(long long _a, long long _b, long long _m = 1000000007) {
  long long _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a) % _m;
    _b /= 2;
    _a = (_a * _a) % _m;
  }
  return _r;
}
bool prime[1000005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  for (long long i = 0; i < 1000005; i++) prime[i] = true;
  for (long long i = 2; i * i <= 1000005; i++) {
    if (prime[i] == true)
      for (long long j = 2 * i; j <= 1000005; j += i) prime[j] = false;
  }
  long long n;
  cin >> n;
  for (long long i = 1; i < 1001; i++) {
    if (prime[n * i + 1] == false) {
      cout << i << endl;
      return 0;
    }
  }
  return 0;
}

```