#include <bits/stdc++.h>
using namespace std;
char a[1010];
int main() {
  int n;
  while (cin >> n) {
    cin >> a;
    if (n % 2 == 0) {
      for (int i = 0; i < n; i++) {
        cout << a[i];
        if (i % 2 == 1 && i != n - 1) cout << "-";
      }
    } else {
      cout << a[0] << a[1] << a[2];
      if (n != 3) cout << "-";
      for (int i = 3; i < n; i++) {
        cout << a[i];
        if (i % 2 == 0 && i != n - 1) {
          cout << "-";
        }
      }
    }
    cout << endl;
  }
}

### Prompt

Your challenge is to write 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;
char a[1010];
int main() {
  int n;
  while (cin >> n) {
    cin >> a;
    if (n % 2 == 0) {
      for (int i = 0; i < n; i++) {
        cout << a[i];
        if (i % 2 == 1 && i != n - 1) cout << "-";
      }
    } else {
      cout << a[0] << a[1] << a[2];
      if (n != 3) cout << "-";
      for (int i = 3; i < n; i++) {
        cout << a[i];
        if (i % 2 == 0 && i != n - 1) {
          cout << "-";
        }
      }
    }
    cout << endl;
  }
}

```