#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  string s;
  cin >> n >> s;
  int p = n % 2, k = 0;
  for (int i = 0; i < s.size() - 1; i++) {
    cout << s[i];
    k++;
    if (k == 2 and p == 0) {
      cout << "-";
      k = 0;
    } else if (k == 3) {
      cout << "-";
      k = 0;
      p = 0;
    }
  }
  cout << s[n - 1];
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 s;
  cin >> n >> s;
  int p = n % 2, k = 0;
  for (int i = 0; i < s.size() - 1; i++) {
    cout << s[i];
    k++;
    if (k == 2 and p == 0) {
      cout << "-";
      k = 0;
    } else if (k == 3) {
      cout << "-";
      k = 0;
      p = 0;
    }
  }
  cout << s[n - 1];
}

```