#include <bits/stdc++.h>
using namespace std;
int main() {
  int num;
  cin >> num;
  char c[100];
  int halat = 0;
  if (num % 3 == 0)
    halat = 3;
  else if (num % 2 == 0)
    halat = 2;
  else if (num % 2 == 1)
    halat = 4;
  if (halat == 4) {
    char c1;
    cin >> c1;
    cout << c1;
    halat = 2;
    num--;
  }
  int count = 0;
  for (int i = 0; i < num; i++) {
    cin >> c[i];
    cout << c[i];
    count++;
    if (count == halat && i < num - 1) {
      count = 0;
      cout << '-';
    }
  }
  cout << endl;
}

### 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;
int main() {
  int num;
  cin >> num;
  char c[100];
  int halat = 0;
  if (num % 3 == 0)
    halat = 3;
  else if (num % 2 == 0)
    halat = 2;
  else if (num % 2 == 1)
    halat = 4;
  if (halat == 4) {
    char c1;
    cin >> c1;
    cout << c1;
    halat = 2;
    num--;
  }
  int count = 0;
  for (int i = 0; i < num; i++) {
    cin >> c[i];
    cout << c[i];
    count++;
    if (count == halat && i < num - 1) {
      count = 0;
      cout << '-';
    }
  }
  cout << endl;
}

```