#include <bits/stdc++.h>
int main() {
  int n(0);
  scanf("%d\n", &n);
  std::string number;
  getline(std::cin, number);
  if (number.size() <= 3) {
    std::cout << number << std::endl;
  } else {
    int index(0);
    if (number.size() % 2) {
      std::cout << number.substr(0, 3);
      index = 3;
    } else {
      std::cout << number.substr(0, 2);
      index = 2;
    }
    while (index < number.size()) {
      std::cout << "-";
      std::cout << number.substr(index, 2);
      index += 2;
    }
    std::cout << std::endl;
  }
  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>
int main() {
  int n(0);
  scanf("%d\n", &n);
  std::string number;
  getline(std::cin, number);
  if (number.size() <= 3) {
    std::cout << number << std::endl;
  } else {
    int index(0);
    if (number.size() % 2) {
      std::cout << number.substr(0, 3);
      index = 3;
    } else {
      std::cout << number.substr(0, 2);
      index = 2;
    }
    while (index < number.size()) {
      std::cout << "-";
      std::cout << number.substr(index, 2);
      index += 2;
    }
    std::cout << std::endl;
  }
  return 0;
}

```