#include <bits/stdc++.h>
using namespace std;
class PhoneNumbers {
 public:
  void solve(std::istream &in, std::ostream &out) {
    int n;
    string s;
    in >> n >> s;
    if (n & 1) {
      int limit = (int)n / 2;
      limit = 2 * (limit - 1);
      for (int i = 0; i < limit; i++) {
        if (i & 1) {
          out << s[i] << '-';
        } else
          out << s[i];
      }
      out << s[n - 3] << s[n - 2] << s[n - 1] << endl;
    } else {
      for (int i = 0; i < n; i++) {
        if ((i & 1) && i != n - 1) {
          out << s[i] << '-';
        } else
          out << s[i];
      }
    }
  }
};
int main() {
  std::ios_base::sync_with_stdio(0);
  std::cin.tie(nullptr);
  PhoneNumbers solver;
  std::istream &in(std::cin);
  std::ostream &out(std::cout);
  solver.solve(in, out);
  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;
class PhoneNumbers {
 public:
  void solve(std::istream &in, std::ostream &out) {
    int n;
    string s;
    in >> n >> s;
    if (n & 1) {
      int limit = (int)n / 2;
      limit = 2 * (limit - 1);
      for (int i = 0; i < limit; i++) {
        if (i & 1) {
          out << s[i] << '-';
        } else
          out << s[i];
      }
      out << s[n - 3] << s[n - 2] << s[n - 1] << endl;
    } else {
      for (int i = 0; i < n; i++) {
        if ((i & 1) && i != n - 1) {
          out << s[i] << '-';
        } else
          out << s[i];
      }
    }
  }
};
int main() {
  std::ios_base::sync_with_stdio(0);
  std::cin.tie(nullptr);
  PhoneNumbers solver;
  std::istream &in(std::cin);
  std::ostream &out(std::cout);
  solver.solve(in, out);
  return 0;
}

```