#include <bits/stdc++.h>
char str[200];
int main(int argc, char *argv[]) {
  int n;
  scanf("%d %s", &n, str);
  if (n & 1) {
    putchar(str[0]);
    putchar(str[1]);
    putchar(str[2]);
    for (n = 3; str[n]; n += 2) {
      putchar('-');
      putchar(str[n]);
      putchar(str[n + 1]);
    }
  } else {
    putchar(str[0]);
    putchar(str[1]);
    for (n = 2; str[n]; n += 2) {
      putchar('-');
      putchar(str[n]);
      putchar(str[n + 1]);
    }
  }
  putchar('\n');
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
char str[200];
int main(int argc, char *argv[]) {
  int n;
  scanf("%d %s", &n, str);
  if (n & 1) {
    putchar(str[0]);
    putchar(str[1]);
    putchar(str[2]);
    for (n = 3; str[n]; n += 2) {
      putchar('-');
      putchar(str[n]);
      putchar(str[n + 1]);
    }
  } else {
    putchar(str[0]);
    putchar(str[1]);
    for (n = 2; str[n]; n += 2) {
      putchar('-');
      putchar(str[n]);
      putchar(str[n + 1]);
    }
  }
  putchar('\n');
}

```