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

### Prompt

Please create a solution in CPP 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 dx[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy[] = {1, -1, 0, 1, -1, 0, 1, -1};
int main() {
  long long int n;
  cin >> n;
  string s;
  cin >> s;
  int k = 0;
  if (n % 2 == 0) {
    for (int i = 0; i < n;) {
      if (k == 2) {
        cout << "-";
        k = 0;
      } else {
        cout << s[i];
        k += 1;
        i++;
      }
    }
  } else {
    if (n <= 3) {
      cout << s;
    } else {
      int i = 0;
      for (i = 0; i < n - 3;) {
        if (k == 2) {
          cout << "-";
          k = 0;
        } else {
          cout << s[i];
          i++;
          k++;
        }
      }
      if (i < n) {
        cout << "-";
        for (; i < n; i++) cout << s[i];
      }
    }
  }
}

```