#include <bits/stdc++.h>
using namespace std;
string x;
int k;
int main() {
  cin >> k >> x;
  if (k <= 3) {
    cout << x << endl;
    return 0;
  }
  for (int i = 0; i < (int)(k / 3 - 1); i++) cout << x.substr(i * 3, 3) << "-";
  if (k % 3 == 2)
    cout << x.substr(k - 5, 3) << "-" << x.substr(k - 2, 2) << endl;
  else if (k % 3 == 1)
    cout << x.substr(k - 4, 2) << "-" << x.substr(k - 2, 2) << endl;
  else
    cout << x.substr(k - 3, 3) << endl;
  return 0;
}

### Prompt

Please formulate 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;
string x;
int k;
int main() {
  cin >> k >> x;
  if (k <= 3) {
    cout << x << endl;
    return 0;
  }
  for (int i = 0; i < (int)(k / 3 - 1); i++) cout << x.substr(i * 3, 3) << "-";
  if (k % 3 == 2)
    cout << x.substr(k - 5, 3) << "-" << x.substr(k - 2, 2) << endl;
  else if (k % 3 == 1)
    cout << x.substr(k - 4, 2) << "-" << x.substr(k - 2, 2) << endl;
  else
    cout << x.substr(k - 3, 3) << endl;
  return 0;
}

```