#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool IsPrime(long long x) {
  bool isPrime = true;
  for (long long a = 2; a * a <= x; a++) {
    if (x % a == 0) return false;
  }
  return true;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  string a;
  cin >> a;
  if (a.length() == 2 || a.length() == 3) {
    cout << a;
    return 0;
  }
  if (a.length() % 2 == 0) {
    for (long long i = 0; i < a.length(); i++) {
      cout << a[i];
      if (i % 2 == 1 && i != n - 1) cout << "-";
    }
  }
  if (a.length() % 2 == 1) {
    for (long long i = 0; i < a.length(); i++) {
      cout << a[i];
      if (i % 2 == 1 && i != n - 2) cout << "-";
    }
  }
}

### Prompt

Please create a solution in cpp to the following problem:

Phone number in Berland is a sequence of n digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given phone number any of its divisions into groups of two or three digits.

Input

The first line contains integer n (2 ≤ n ≤ 100) — amount of digits in the phone number. The second line contains n digits — the phone number to divide into groups.

Output

Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any.

Examples

Input

6
549871


Output

54-98-71

Input

7
1198733


Output

11-987-33

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool IsPrime(long long x) {
  bool isPrime = true;
  for (long long a = 2; a * a <= x; a++) {
    if (x % a == 0) return false;
  }
  return true;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  string a;
  cin >> a;
  if (a.length() == 2 || a.length() == 3) {
    cout << a;
    return 0;
  }
  if (a.length() % 2 == 0) {
    for (long long i = 0; i < a.length(); i++) {
      cout << a[i];
      if (i % 2 == 1 && i != n - 1) cout << "-";
    }
  }
  if (a.length() % 2 == 1) {
    for (long long i = 0; i < a.length(); i++) {
      cout << a[i];
      if (i % 2 == 1 && i != n - 2) cout << "-";
    }
  }
}

```