#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i, j, two, three;
  char c[110];
  cin >> n;
  while (cin.fail() == false) {
    cin >> c;
    for (i = 0; i < 51; i++) {
      for (j = 0; j < 35; j++) {
        if (i * 2 + j * 3 == n) {
          two = i;
          three = j;
          i = 100;
          j = 100;
          break;
        }
      }
    }
    int k = 0;
    for (i = 0; i < two; i++) {
      cout << c[k] << c[k + 1];
      k += 2;
      if (k < n - 1) {
        cout << "-";
      }
    }
    for (i = 0; i < three; i++) {
      cout << c[k] << c[k + 1] << c[k + 2];
      k += 3;
      if (k < n - 1) {
        cout << "-";
      }
    }
    cout << endl;
    cin >> n;
  }
  return 0;
}

### Prompt

Your task is to create 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 n, i, j, two, three;
  char c[110];
  cin >> n;
  while (cin.fail() == false) {
    cin >> c;
    for (i = 0; i < 51; i++) {
      for (j = 0; j < 35; j++) {
        if (i * 2 + j * 3 == n) {
          two = i;
          three = j;
          i = 100;
          j = 100;
          break;
        }
      }
    }
    int k = 0;
    for (i = 0; i < two; i++) {
      cout << c[k] << c[k + 1];
      k += 2;
      if (k < n - 1) {
        cout << "-";
      }
    }
    for (i = 0; i < three; i++) {
      cout << c[k] << c[k + 1] << c[k + 2];
      k += 3;
      if (k < n - 1) {
        cout << "-";
      }
    }
    cout << endl;
    cin >> n;
  }
  return 0;
}

```