#include <bits/stdc++.h>
using namespace std;
int n;
char s[200];
int main() {
  cin >> n;
  scanf("%s", s);
  if (n % 2 == 0) {
    for (int i = 0; i < (int)(n - 1); i++) {
      printf("%c", s[i]);
      if (i % 2) printf("-");
    }
    printf("%c", s[n - 1]);
  } else {
    for (int i = 0; i < (int)(n - 2); i++) {
      printf("%c", s[i]);
      if (i % 2) printf("-");
    }
    printf("%c%c", s[n - 2], s[n - 1]);
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 n;
char s[200];
int main() {
  cin >> n;
  scanf("%s", s);
  if (n % 2 == 0) {
    for (int i = 0; i < (int)(n - 1); i++) {
      printf("%c", s[i]);
      if (i % 2) printf("-");
    }
    printf("%c", s[n - 1]);
  } else {
    for (int i = 0; i < (int)(n - 2); i++) {
      printf("%c", s[i]);
      if (i % 2) printf("-");
    }
    printf("%c%c", s[n - 2], s[n - 1]);
  }
}

```