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

### 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, j;
  scanf("%d", &n);
  char a[n + 1];
  scanf("%s", a);
  if (n % 2 == 0) {
    printf("%c%c", a[0], a[1]);
    for (i = 2; a[i] != '\0'; i += 2) {
      printf("-");
      for (j = i; j < i + 2; j++) printf("%c", a[j]);
    }
  } else {
    printf("%c%c%c", a[0], a[1], a[2]);
    for (i = 3; a[i] != '\0'; i += 2) {
      printf("-");
      for (j = i; j < i + 2; j++) printf("%c", a[j]);
    }
  }
}

```