#include <bits/stdc++.h>
int main() {
  int n;
  char s[105];
  scanf("%d", &n);
  scanf("%s", s);
  int d = n % 2;
  for (int i = 0; i < n; i += 2) {
    if (i) printf("-");
    if (d > 0)
      printf("%c%c%c", s[i++], s[i + 1], s[i + 2]);
    else
      printf("%c%c", s[i], s[i + 1]);
    d--;
  }
}

### Prompt

Please create a solution in CPP 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;
  char s[105];
  scanf("%d", &n);
  scanf("%s", s);
  int d = n % 2;
  for (int i = 0; i < n; i += 2) {
    if (i) printf("-");
    if (d > 0)
      printf("%c%c%c", s[i++], s[i + 1], s[i + 2]);
    else
      printf("%c%c", s[i], s[i + 1]);
    d--;
  }
}

```