#include <bits/stdc++.h>
using namespace std;
int a[10000];
int main() {
  int n;
  int flag;
  char s[1000];
  while (~scanf("%d", &n)) {
    getchar();
    gets(s);
    int len = strlen(s);
    if (len % 2 == 0) {
      for (int i = 0; i < len; i++) {
        if ((i + 1) % 2 == 0 && i != len - 1)
          printf("%c-", s[i]);
        else
          printf("%c", s[i]);
      }
      printf("\n");
      continue;
    } else {
      for (int i = 0; i < len - 3; i++) {
        if ((i + 1) % 2 == 0 && i != len - 1)
          printf("%c-", s[i]);
        else
          printf("%c", s[i]);
      }
      for (int i = len - 3; i < len; i++) printf("%c", s[i]);
      printf("\n");
      continue;
    }
  }
  return 0;
}

### 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 a[10000];
int main() {
  int n;
  int flag;
  char s[1000];
  while (~scanf("%d", &n)) {
    getchar();
    gets(s);
    int len = strlen(s);
    if (len % 2 == 0) {
      for (int i = 0; i < len; i++) {
        if ((i + 1) % 2 == 0 && i != len - 1)
          printf("%c-", s[i]);
        else
          printf("%c", s[i]);
      }
      printf("\n");
      continue;
    } else {
      for (int i = 0; i < len - 3; i++) {
        if ((i + 1) % 2 == 0 && i != len - 1)
          printf("%c-", s[i]);
        else
          printf("%c", s[i]);
      }
      for (int i = len - 3; i < len; i++) printf("%c", s[i]);
      printf("\n");
      continue;
    }
  }
  return 0;
}

```