#include <bits/stdc++.h>
using namespace std;
int main() {
  long n, k, i;
  string s;
  cin >> k;
  cin >> s;
  if (k % 2 == 0)
    for (i = 0; i < k; i = i + 2)
      if (i != (k - 2))
        cout << s[i] << s[i + 1] << '-';
      else
        cout << s[i] << s[i + 1];
  else {
    if (k != 3) {
      for (i = 0; i < k - 3; i = i + 2) cout << s[i] << s[i + 1] << '-';
      for (i = k - 3; i < k; i++) cout << s[i];
    } else
      for (i = 0; i < k; i++) cout << s[i];
  }
  return 0;
}

### 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;
int main() {
  long n, k, i;
  string s;
  cin >> k;
  cin >> s;
  if (k % 2 == 0)
    for (i = 0; i < k; i = i + 2)
      if (i != (k - 2))
        cout << s[i] << s[i + 1] << '-';
      else
        cout << s[i] << s[i + 1];
  else {
    if (k != 3) {
      for (i = 0; i < k - 3; i = i + 2) cout << s[i] << s[i + 1] << '-';
      for (i = k - 3; i < k; i++) cout << s[i];
    } else
      for (i = 0; i < k; i++) cout << s[i];
  }
  return 0;
}

```