#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& cout, const vector<T>& a) {
  if ((int)a.size() == 0) {
    return (cout << "()");
  }
  cout << "(" << a[0];
  for (int i = 1; i < (int)a.size(); i++) {
    cout << "; " << a[i];
  }
  return (cout << ")");
}
template <typename T>
ostream& operator<<(ostream& cout, const set<T>& a) {
  if ((int)a.size() == 0) {
    return (cout << "{}");
  }
  auto it = a.begin();
  cout << "{" << *it;
  ++it;
  for (; it != a.end(); ++it) {
    cout << "; " << *it;
  }
  return (cout << "}");
}
template <typename T>
ostream& operator<<(ostream& cout, const multiset<T>& a) {
  if ((int)a.size() == 0) {
    return (cout << "{}");
  }
  auto it = a.begin();
  cout << "{" << *it;
  ++it;
  for (; it != a.end(); ++it) {
    cout << "; " << *it;
  }
  return (cout << "}");
}
template <typename T1, typename T2>
ostream& operator<<(ostream& cout, const pair<T1, T2>& a) {
  return cout << "(" << a.first << "; " << a.second << ")";
}
const int nmax = 2000 * 1000;
const int inf = 2000 * 1000 * 1000;
const long long infl = 1000ll * 1000ll * 1000ll * 1000ll * 1000ll * 1000ll;
const int mod = 1000 * 1000 * 1000 + 7;
const long double pi = acos(-1.0);
int n;
bool used[nmax];
int main() {
  cin >> n;
  for (long long i = 2; i < nmax; i++) {
    if (used[i]) {
      continue;
    }
    for (long long j = i * i; j < nmax; j += i) {
      used[j] = true;
    }
  }
  for (int i = 1; i <= 1000; i++) {
    if (used[i * n + 1]) {
      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;
template <typename T>
ostream& operator<<(ostream& cout, const vector<T>& a) {
  if ((int)a.size() == 0) {
    return (cout << "()");
  }
  cout << "(" << a[0];
  for (int i = 1; i < (int)a.size(); i++) {
    cout << "; " << a[i];
  }
  return (cout << ")");
}
template <typename T>
ostream& operator<<(ostream& cout, const set<T>& a) {
  if ((int)a.size() == 0) {
    return (cout << "{}");
  }
  auto it = a.begin();
  cout << "{" << *it;
  ++it;
  for (; it != a.end(); ++it) {
    cout << "; " << *it;
  }
  return (cout << "}");
}
template <typename T>
ostream& operator<<(ostream& cout, const multiset<T>& a) {
  if ((int)a.size() == 0) {
    return (cout << "{}");
  }
  auto it = a.begin();
  cout << "{" << *it;
  ++it;
  for (; it != a.end(); ++it) {
    cout << "; " << *it;
  }
  return (cout << "}");
}
template <typename T1, typename T2>
ostream& operator<<(ostream& cout, const pair<T1, T2>& a) {
  return cout << "(" << a.first << "; " << a.second << ")";
}
const int nmax = 2000 * 1000;
const int inf = 2000 * 1000 * 1000;
const long long infl = 1000ll * 1000ll * 1000ll * 1000ll * 1000ll * 1000ll;
const int mod = 1000 * 1000 * 1000 + 7;
const long double pi = acos(-1.0);
int n;
bool used[nmax];
int main() {
  cin >> n;
  for (long long i = 2; i < nmax; i++) {
    if (used[i]) {
      continue;
    }
    for (long long j = i * i; j < nmax; j += i) {
      used[j] = true;
    }
  }
  for (int i = 1; i <= 1000; i++) {
    if (used[i * n + 1]) {
      cout << i << endl;
      return 0;
    }
  }
  return 0;
}

```