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

### Prompt

Develop a solution in CPP to the problem described below:

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 len;
  string s;
  cin >> len;
  cin >> s;
  if (len % 2 == 1) {
    for (int i = 0; i < len - 3; i++)
      if (i % 2 == 1)
        cout << s[i] << "-";
      else
        cout << s[i];
    ;
    cout << s.substr(len - 3, 3) << '\n';
  } else {
    for (int i = 0; i < len - 2; i++)
      if (i % 2 == 1)
        cout << s[i] << "-";
      else
        cout << s[i];
    cout << s.substr(len - 2, 2) << '\n';
  }
  return 0;
}

```