#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  string ss;
  cin >> n >> ss;
  if (n % 2 == 0) {
    for (int i = 0, k = 1; i < ss.size(); ++i) {
      cout << ss[i];
      if (k % 2 == 0 && i < ss.size() - 2) {
        cout << "-";
      }
      ++k;
    }
    cout << endl;
  } else if (n % 3 == 0) {
    for (int i = 0, k = 1; i < ss.size(); ++i) {
      cout << ss[i];
      if (k % 3 == 0 && i < ss.size() - 2) {
        cout << "-";
      }
      ++k;
    }
    cout << endl;
  } else {
    cout << ss[0] << ss[1] << ss[2] << "-";
    for (int i = 3, k = 1; i < ss.size(); ++i) {
      cout << ss[i];
      if (k % 2 == 0 && i < ss.size() - 2) {
        cout << "-";
      }
      ++k;
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  string ss;
  cin >> n >> ss;
  if (n % 2 == 0) {
    for (int i = 0, k = 1; i < ss.size(); ++i) {
      cout << ss[i];
      if (k % 2 == 0 && i < ss.size() - 2) {
        cout << "-";
      }
      ++k;
    }
    cout << endl;
  } else if (n % 3 == 0) {
    for (int i = 0, k = 1; i < ss.size(); ++i) {
      cout << ss[i];
      if (k % 3 == 0 && i < ss.size() - 2) {
        cout << "-";
      }
      ++k;
    }
    cout << endl;
  } else {
    cout << ss[0] << ss[1] << ss[2] << "-";
    for (int i = 3, k = 1; i < ss.size(); ++i) {
      cout << ss[i];
      if (k % 2 == 0 && i < ss.size() - 2) {
        cout << "-";
      }
      ++k;
    }
    cout << endl;
  }
  return 0;
}

```