#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  char str[110];
  scanf("%d", &n);
  scanf("%s", str);
  if (n % 3 == 0) {
    for (int i = 0; i < n; i += 3) {
      if (i) putchar('-');
      putchar(str[i]);
      putchar(str[i + 1]);
      putchar(str[i + 2]);
    }
  } else if (n % 3 == 1) {
    putchar(str[0]);
    putchar(str[1]);
    putchar('-');
    putchar(str[2]);
    putchar(str[3]);
    for (int i = 4; i < n; i += 3) {
      putchar('-');
      putchar(str[i]);
      putchar(str[i + 1]);
      putchar(str[i + 2]);
    }
  } else {
    putchar(str[0]);
    putchar(str[1]);
    for (int i = 2; i < n; i += 3) {
      putchar('-');
      putchar(str[i]);
      putchar(str[i + 1]);
      putchar(str[i + 2]);
    }
  }
  printf("\n");
  return 0;
}

### Prompt

Create a solution in cpp for 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;
int main() {
  int n;
  char str[110];
  scanf("%d", &n);
  scanf("%s", str);
  if (n % 3 == 0) {
    for (int i = 0; i < n; i += 3) {
      if (i) putchar('-');
      putchar(str[i]);
      putchar(str[i + 1]);
      putchar(str[i + 2]);
    }
  } else if (n % 3 == 1) {
    putchar(str[0]);
    putchar(str[1]);
    putchar('-');
    putchar(str[2]);
    putchar(str[3]);
    for (int i = 4; i < n; i += 3) {
      putchar('-');
      putchar(str[i]);
      putchar(str[i + 1]);
      putchar(str[i + 2]);
    }
  } else {
    putchar(str[0]);
    putchar(str[1]);
    for (int i = 2; i < n; i += 3) {
      putchar('-');
      putchar(str[i]);
      putchar(str[i + 1]);
      putchar(str[i + 2]);
    }
  }
  printf("\n");
  return 0;
}

```