#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i, j;
  char str[105];
  cin >> n;
  cin >> str;
  i = 0;
  while (i < n - 1) {
    if ((n - 1) - (i) == 2) {
      cout << str[i] << str[i + 1] << str[i + 2];
      i = i + 3;
    } else {
      cout << str[i] << str[i + 1];
      i = i + 2;
    }
    if (i < n - 1) cout << "-";
  }
  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 n, i, j;
  char str[105];
  cin >> n;
  cin >> str;
  i = 0;
  while (i < n - 1) {
    if ((n - 1) - (i) == 2) {
      cout << str[i] << str[i + 1] << str[i + 2];
      i = i + 3;
    } else {
      cout << str[i] << str[i + 1];
      i = i + 2;
    }
    if (i < n - 1) cout << "-";
  }
  return 0;
}

```