#include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000;
const long long LINF = 1e17;
const double PI = 3.141592653589793238;
const int mod = 1000000007;
const double eps = 0.000001;
template <typename T, typename TT>
ostream &operator<<(ostream &s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream &operator<<(ostream &s, vector<T> t) {
  for (int i = 0; i < (t).size(); i++)
    s << t[i] << ((i < (t).size() - 1) ? " " : "");
  return s;
}
template <typename T>
ostream &operator<<(ostream &s, set<T> t) {
  for (T x : t) s << x << " ";
  return s;
}
template <typename T>
istream &operator>>(istream &s, vector<T> &t) {
  for (int _i = 0; _i < t.size(); _i++) s >> t[_i];
  return s;
}
int isprime(int n) {
  if (n == 1) return 0;
  if (n == 2) return 1;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int main(int argc, const char *argv[]) {
  int n;
  cin >> n;
  for (int m = (1); m < (1000); m++) {
    if (!isprime(n * m + 1)) {
      cout << m << "\n";
      break;
    }
  }
}

### Prompt

Generate 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 int INF = 1000000000;
const long long LINF = 1e17;
const double PI = 3.141592653589793238;
const int mod = 1000000007;
const double eps = 0.000001;
template <typename T, typename TT>
ostream &operator<<(ostream &s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream &operator<<(ostream &s, vector<T> t) {
  for (int i = 0; i < (t).size(); i++)
    s << t[i] << ((i < (t).size() - 1) ? " " : "");
  return s;
}
template <typename T>
ostream &operator<<(ostream &s, set<T> t) {
  for (T x : t) s << x << " ";
  return s;
}
template <typename T>
istream &operator>>(istream &s, vector<T> &t) {
  for (int _i = 0; _i < t.size(); _i++) s >> t[_i];
  return s;
}
int isprime(int n) {
  if (n == 1) return 0;
  if (n == 2) return 1;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int main(int argc, const char *argv[]) {
  int n;
  cin >> n;
  for (int m = (1); m < (1000); m++) {
    if (!isprime(n * m + 1)) {
      cout << m << "\n";
      break;
    }
  }
}

```