#include <bits/stdc++.h>
int main() {
  int n, i, x = 0;
  int s1[300];
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%1d", &s1[i]);
  }
  if (n % 2 == 0) {
    for (i = 1; i <= n; i++) {
      printf("%d", s1[i]);
      x++;
      if (x % 2 == 0 && i != n) {
        printf("-");
      }
    }
  }
  if (n % 2 != 0) {
    for (i = 1; i <= n; i++) {
      if (i != n - 2) {
        printf("%d", s1[i]);
        x++;
        if (x % 2 == 0 && i != n) {
          printf("-");
        }
      } else {
        printf("%d", s1[i]);
      }
    }
  }
}

### Prompt

Your task is to create a CPP solution 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, i, x = 0;
  int s1[300];
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%1d", &s1[i]);
  }
  if (n % 2 == 0) {
    for (i = 1; i <= n; i++) {
      printf("%d", s1[i]);
      x++;
      if (x % 2 == 0 && i != n) {
        printf("-");
      }
    }
  }
  if (n % 2 != 0) {
    for (i = 1; i <= n; i++) {
      if (i != n - 2) {
        printf("%d", s1[i]);
        x++;
        if (x % 2 == 0 && i != n) {
          printf("-");
        }
      } else {
        printf("%d", s1[i]);
      }
    }
  }
}

```