#include <bits/stdc++.h>
using namespace std;
int n;
char s[110];
int main() {
  scanf("%d%s", &n, s);
  for (int i = 0; i < n; i += 2)
    if (n - i <= 3) {
      printf("%s\n", s + i);
      break;
    } else
      printf("%c%c-", s[i], s[i + 1]);
  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;
int n;
char s[110];
int main() {
  scanf("%d%s", &n, s);
  for (int i = 0; i < n; i += 2)
    if (n - i <= 3) {
      printf("%s\n", s + i);
      break;
    } else
      printf("%c%c-", s[i], s[i + 1]);
  return 0;
}

```