#include <bits/stdc++.h>
using namespace std;
int main(void) {
  int n, limit, i;
  cin >> n;
  char x[101];
  cin >> x;
  limit = (n & 1) ? n - 3 : n - 2;
  for (i = 0; i < limit; i += 2) cout << x[i] << x[i + 1] << '-';
  for (; i < n; ++i) cout << x[i];
  cout << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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(void) {
  int n, limit, i;
  cin >> n;
  char x[101];
  cin >> x;
  limit = (n & 1) ? n - 3 : n - 2;
  for (i = 0; i < limit; i += 2) cout << x[i] << x[i + 1] << '-';
  for (; i < n; ++i) cout << x[i];
  cout << endl;
  return 0;
}

```