#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("-ffloat-store")
const long long INF = 0xFFFFFFFFFFFFFFFL;
const long long N = 1e9 + 7;
vector<long long> pri;
void yo(int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) pri.push_back(p);
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i < 1001; i++) {
    if (!binary_search((pri).begin(), (pri).end(), n * i + 1)) {
      cout << i;
      return;
    }
  }
}
int32_t main(int32_t argc, char **argv) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  yo(1000003);
  long long t;
  t = 1;
  while (t--) solve();
  cerr << "\n"
       << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("-ffloat-store")
const long long INF = 0xFFFFFFFFFFFFFFFL;
const long long N = 1e9 + 7;
vector<long long> pri;
void yo(int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) pri.push_back(p);
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i < 1001; i++) {
    if (!binary_search((pri).begin(), (pri).end(), n * i + 1)) {
      cout << i;
      return;
    }
  }
}
int32_t main(int32_t argc, char **argv) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  yo(1000003);
  long long t;
  t = 1;
  while (t--) solve();
  cerr << "\n"
       << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}

```