#include <bits/stdc++.h>
int main() {
  int n, i;
  char s[101];
  scanf("%d", &n);
  scanf("%s", s);
  printf("%c%c", s[0], s[1]);
  for (i = 2; i <= n - 2; i += 2) {
    printf("-%c%c", s[i], s[i + 1]);
  }
  if (i < n) printf("%c", s[i]);
  return 0;
}

### Prompt

Your task is to create 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>
int main() {
  int n, i;
  char s[101];
  scanf("%d", &n);
  scanf("%s", s);
  printf("%c%c", s[0], s[1]);
  for (i = 2; i <= n - 2; i += 2) {
    printf("-%c%c", s[i], s[i + 1]);
  }
  if (i < n) printf("%c", s[i]);
  return 0;
}

```