#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  char arr[200];
  cin >> arr;
  if (n <= 3)
    cout << arr;
  else {
    if (n % 2 == 0) {
      for (int i = 1; i <= n; i++) {
        cout << arr[i - 1];
        if (i % 2 == 0 && i != n) cout << "-";
      }
    } else {
      for (int i = 1; i <= n; i++) {
        cout << arr[i - 1];
        if (i % 2 == 0 && i != n - 1) cout << "-";
      }
    }
  }
}

### Prompt

Generate 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;
  cin >> n;
  char arr[200];
  cin >> arr;
  if (n <= 3)
    cout << arr;
  else {
    if (n % 2 == 0) {
      for (int i = 1; i <= n; i++) {
        cout << arr[i - 1];
        if (i % 2 == 0 && i != n) cout << "-";
      }
    } else {
      for (int i = 1; i <= n; i++) {
        cout << arr[i - 1];
        if (i % 2 == 0 && i != n - 1) cout << "-";
      }
    }
  }
}

```