#include <bits/stdc++.h>
int n;
char num[10001];
int main() {
  std::cin >> n;
  for (int i = 1; i <= n; i++) std::cin >> num[i];
  for (int i = 1; i <= n - (n % 2 ? 3 : 0); i++) {
    std::cout << num[i];
    if (i % 2 == 0 && i != n) std::cout << '-';
  }
  if (n % 2) std::cout << num[n - 2] << num[n - 1] << num[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>
int n;
char num[10001];
int main() {
  std::cin >> n;
  for (int i = 1; i <= n; i++) std::cin >> num[i];
  for (int i = 1; i <= n - (n % 2 ? 3 : 0); i++) {
    std::cout << num[i];
    if (i % 2 == 0 && i != n) std::cout << '-';
  }
  if (n % 2) std::cout << num[n - 2] << num[n - 1] << num[n];
  return 0;
}

```