#include <bits/stdc++.h>
using namespace std;
int main() {
  int y;
  string x;
  cin >> y >> x;
  if (y <= 3)
    cout << x;
  else {
    if (y % 2) {
      cout << x[0] << x[1] << x[2] << '-';
      x.erase(0, 3);
    } else {
      cout << x[0] << x[1] << '-';
      x.erase(0, 2);
    }
    while (x.size() > 2) {
      cout << x[0] << x[1] << '-';
      x.erase(0, 2);
    }
    cout << x[0] << x[1];
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 y;
  string x;
  cin >> y >> x;
  if (y <= 3)
    cout << x;
  else {
    if (y % 2) {
      cout << x[0] << x[1] << x[2] << '-';
      x.erase(0, 3);
    } else {
      cout << x[0] << x[1] << '-';
      x.erase(0, 2);
    }
    while (x.size() > 2) {
      cout << x[0] << x[1] << '-';
      x.erase(0, 2);
    }
    cout << x[0] << x[1];
  }
  return 0;
}

```