#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  char num[120];
  scanf("%d", &n);
  scanf("%s", num);
  if (n == 4) {
    printf("%c%c%c%c%c", num[0], num[1], '-', num[2], num[3]);
    return 0;
  }
  for (int i = 0, j = 0; i < n; i++, j++) {
    printf("%c", num[i]);
    if (j == 2 && i < n - 1 && n % 2 != 0) {
      printf("-");
      j = 0;
    } else if (j == 2 && i < n - 2 && n % 2 == 0) {
      printf("-");
      j = 0;
    }
  }
}

### 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>
using namespace std;
int main() {
  int n;
  char num[120];
  scanf("%d", &n);
  scanf("%s", num);
  if (n == 4) {
    printf("%c%c%c%c%c", num[0], num[1], '-', num[2], num[3]);
    return 0;
  }
  for (int i = 0, j = 0; i < n; i++, j++) {
    printf("%c", num[i]);
    if (j == 2 && i < n - 1 && n % 2 != 0) {
      printf("-");
      j = 0;
    } else if (j == 2 && i < n - 2 && n % 2 == 0) {
      printf("-");
      j = 0;
    }
  }
}

```