#include <bits/stdc++.h>
using namespace std;
int number;
string input, str, output;
vector<string> vec;
void odd_number() {
  output = input.substr(0, 3) + "-";
  for (int i = 3; i < number; i += 2) {
    str = input.substr(i, 2);
    vec.push_back(str);
    output += vec.back() + "-";
  }
  output = output.substr(0, output.length() - 1);
  cout << output;
}
void even_number() {
  for (int i = 0; i < number; i += 2) {
    str = input.substr(i, 2);
    vec.push_back(str);
    output += vec.back() + "-";
  }
  output = output.substr(0, output.length() - 1);
  cout << output;
}
void judge() {
  if (number % 2 == 0) {
    even_number();
  } else {
    odd_number();
  }
}
int main(int argc, char** argv) {
  cin >> number >> input;
  judge();
}

### 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 number;
string input, str, output;
vector<string> vec;
void odd_number() {
  output = input.substr(0, 3) + "-";
  for (int i = 3; i < number; i += 2) {
    str = input.substr(i, 2);
    vec.push_back(str);
    output += vec.back() + "-";
  }
  output = output.substr(0, output.length() - 1);
  cout << output;
}
void even_number() {
  for (int i = 0; i < number; i += 2) {
    str = input.substr(i, 2);
    vec.push_back(str);
    output += vec.back() + "-";
  }
  output = output.substr(0, output.length() - 1);
  cout << output;
}
void judge() {
  if (number % 2 == 0) {
    even_number();
  } else {
    odd_number();
  }
}
int main(int argc, char** argv) {
  cin >> number >> input;
  judge();
}

```