#include <bits/stdc++.h>
using namespace std;
long long a[1000001], n, m = INT_MIN, b[1000001], c, k1, c1, r, l, x,
                         m1 = INT_MAX, p1, p;
double qx;
string s, s1, s2;
char q;
set<char> w, e;
bool u[1000001];
pair<long long, long long> d[100001];
long long f(int c) {
  int k = 0;
  for (int i = 2; i <= sqrt(c); i++) {
    if (c % i == 0) {
      k++;
      r = i;
      break;
    }
  }
  if (k == 1)
    return true;
  else
    return false;
}
int main() {
  cin >> n;
  m = 1;
  while (1) {
    c = n * m + 1;
    f(c);
    if (f(c)) return cout << m, 0;
    m++;
  }
}

### Prompt

In cpp, your task is to solve 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;
long long a[1000001], n, m = INT_MIN, b[1000001], c, k1, c1, r, l, x,
                         m1 = INT_MAX, p1, p;
double qx;
string s, s1, s2;
char q;
set<char> w, e;
bool u[1000001];
pair<long long, long long> d[100001];
long long f(int c) {
  int k = 0;
  for (int i = 2; i <= sqrt(c); i++) {
    if (c % i == 0) {
      k++;
      r = i;
      break;
    }
  }
  if (k == 1)
    return true;
  else
    return false;
}
int main() {
  cin >> n;
  m = 1;
  while (1) {
    c = n * m + 1;
    f(c);
    if (f(c)) return cout << m, 0;
    m++;
  }
}

```