#include <bits/stdc++.h>
using namespace std;
const long long int INF = (long long int)LLONG_MAX;
int dx[4] = {0, -1, 0, 1};
int dy[4] = {1, 0, -1, 0};
int dx8[9] = {1, 0, -1, 0, 1, -1, -1, 1, 0};
int dy8[9] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
long long int ipow(long long int base, long long int exp) {
  long long int result = 1;
  while (exp) {
    if (exp & 1) result = (result * base);
    exp >>= 1;
    base = (base * base);
  }
  return result;
}
string tos(long long int n) {
  stringstream ss;
  string ans;
  ss << n;
  ss >> ans;
  return ans;
}
long long int toll(string n) { return atoll(n.c_str()); }
long long int dist(long long int x1, long long int y1, long long int x2,
                   long long int y2) {
  return ((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2));
}
int doubleCompare(double x, double y) {
  if (fabs(x - y) <= 1e-100) return 0;
  if (x < y) return -1;
  return 1;
}
long long int mod(long long int x, long long int m) { return (x % m + m) % m; }
template <typename T>
T modpow(T base, T exp, T modulus) {
  base %= modulus;
  T result = 1;
  while (exp > 0) {
    if (exp & 1) result = (result * base) % modulus;
    base = (base * base) % modulus;
    exp >>= 1;
  }
  return result;
}
bool arr[100000];
void seive() {
  arr[0] = arr[1] = 1;
  for (int i = 2; i < 100000; i++) {
    if (!arr[i]) {
      for (int j = i * 2; j < 100000; j += i) {
        arr[j] = 1;
      }
    }
  }
}
int main() {
  memset(arr, 0, sizeof(arr));
  int n;
  cin >> n;
  seive();
  for (int i = 1; i <= 1000; i++) {
    if (n * i + 1 < 100000 && arr[n * i + 1]) {
      cout << i << "\n";
      return 0;
    }
  }
}

### Prompt

Create a solution in cpp for 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 long long int INF = (long long int)LLONG_MAX;
int dx[4] = {0, -1, 0, 1};
int dy[4] = {1, 0, -1, 0};
int dx8[9] = {1, 0, -1, 0, 1, -1, -1, 1, 0};
int dy8[9] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
long long int ipow(long long int base, long long int exp) {
  long long int result = 1;
  while (exp) {
    if (exp & 1) result = (result * base);
    exp >>= 1;
    base = (base * base);
  }
  return result;
}
string tos(long long int n) {
  stringstream ss;
  string ans;
  ss << n;
  ss >> ans;
  return ans;
}
long long int toll(string n) { return atoll(n.c_str()); }
long long int dist(long long int x1, long long int y1, long long int x2,
                   long long int y2) {
  return ((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2));
}
int doubleCompare(double x, double y) {
  if (fabs(x - y) <= 1e-100) return 0;
  if (x < y) return -1;
  return 1;
}
long long int mod(long long int x, long long int m) { return (x % m + m) % m; }
template <typename T>
T modpow(T base, T exp, T modulus) {
  base %= modulus;
  T result = 1;
  while (exp > 0) {
    if (exp & 1) result = (result * base) % modulus;
    base = (base * base) % modulus;
    exp >>= 1;
  }
  return result;
}
bool arr[100000];
void seive() {
  arr[0] = arr[1] = 1;
  for (int i = 2; i < 100000; i++) {
    if (!arr[i]) {
      for (int j = i * 2; j < 100000; j += i) {
        arr[j] = 1;
      }
    }
  }
}
int main() {
  memset(arr, 0, sizeof(arr));
  int n;
  cin >> n;
  seive();
  for (int i = 1; i <= 1000; i++) {
    if (n * i + 1 < 100000 && arr[n * i + 1]) {
      cout << i << "\n";
      return 0;
    }
  }
}

```