#include <bits/stdc++.h>
using namespace std;
int n;
string s;
int main() {
  scanf("%d", &n);
  cin >> s;
  if (n <= 3)
    cout << s;
  else {
    if (n % 2) {
      cout << s.substr(0, 3) << "-";
      for (int i = 3; i < s.length(); i++) {
        printf("%c", s[i]);
        if (i % 2 == 0 && i != s.length() - 1) printf("-");
      }
    } else
      for (int i = 0; i < s.length(); i++) {
        printf("%c", s[i]);
        if (i % 2 == 1 && i != s.length() - 1) printf("-");
      }
  }
  printf("\n");
  return 0;
}

### Prompt

Create a solution in CPP for 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 n;
string s;
int main() {
  scanf("%d", &n);
  cin >> s;
  if (n <= 3)
    cout << s;
  else {
    if (n % 2) {
      cout << s.substr(0, 3) << "-";
      for (int i = 3; i < s.length(); i++) {
        printf("%c", s[i]);
        if (i % 2 == 0 && i != s.length() - 1) printf("-");
      }
    } else
      for (int i = 0; i < s.length(); i++) {
        printf("%c", s[i]);
        if (i % 2 == 1 && i != s.length() - 1) printf("-");
      }
  }
  printf("\n");
  return 0;
}

```