#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  string m;
  cin >> n;
  cin >> m;
  if (m.size() % 2 == 0)
    for (int i = 0; i < n; i++) {
      cout << m[i];
      if (i % 2 != 0 && i != n - 1) cout << "-";
    }
  else {
    if (n >= 5) {
      cout << m[0] << m[1] << "-" << m[2] << m[3] << m[4];
      if (n - 5 != 0) {
        cout << "-";
        for (int i = 5; i < n; i++) {
          cout << m[i];
          if (i % 2 == 0 && i != n - 1) cout << "-";
        }
      }
    } else
      cout << m;
  }
}

### Prompt

Your task is to create a Cpp solution 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;
int main() {
  int n;
  string m;
  cin >> n;
  cin >> m;
  if (m.size() % 2 == 0)
    for (int i = 0; i < n; i++) {
      cout << m[i];
      if (i % 2 != 0 && i != n - 1) cout << "-";
    }
  else {
    if (n >= 5) {
      cout << m[0] << m[1] << "-" << m[2] << m[3] << m[4];
      if (n - 5 != 0) {
        cout << "-";
        for (int i = 5; i < n; i++) {
          cout << m[i];
          if (i % 2 == 0 && i != n - 1) cout << "-";
        }
      }
    } else
      cout << m;
  }
}

```