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

### Prompt

Generate 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>
using namespace std;
int main() {
  int n, i = 0;
  string str;
  scanf("%d", &n);
  cin >> str;
  if (n % 2 == 0) {
    i = 2;
    printf("%c%c", str[0], str[1]);
  } else {
    i = 3;
    printf("%c%c%c", str[0], str[1], str[2]);
  }
  for (; i < n; i += 2) {
    printf("-%c%c", str[i], str[i + 1]);
  }
  return 0;
}

```