#include <bits/stdc++.h>
using namespace std;
char a;
int main() {
  int n, i;
  while (scanf("%d", &n) != EOF) {
    getchar();
    for (i = 1; i <= n; i++) {
      a = getchar();
      printf("%c", a);
      if (i % 2 == 0 && ((n - i) > 2 || (n - i > 1))) printf("-");
    }
    printf("\n");
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
char a;
int main() {
  int n, i;
  while (scanf("%d", &n) != EOF) {
    getchar();
    for (i = 1; i <= n; i++) {
      a = getchar();
      printf("%c", a);
      if (i % 2 == 0 && ((n - i) > 2 || (n - i > 1))) printf("-");
    }
    printf("\n");
  }
  return 0;
}

```