#include <bits/stdc++.h>
using namespace std;
int n;
char s[101];
int main() {
  scanf("%d%s", &n, s);
  if (n <= 3) {
    puts(s);
    return 0;
  }
  if (n & 1) {
    printf("%c%c%c", s[0], s[1], s[2]);
    for (int i = 3; i < n; i += 2) printf("-%c%c", s[i], s[i + 1]);
    puts("");
  } else {
    for (int i = 0; i < n; i += 2)
      printf("%s%c%c", i ? "-" : "", s[i], s[i + 1]);
    puts("");
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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>
using namespace std;
int n;
char s[101];
int main() {
  scanf("%d%s", &n, s);
  if (n <= 3) {
    puts(s);
    return 0;
  }
  if (n & 1) {
    printf("%c%c%c", s[0], s[1], s[2]);
    for (int i = 3; i < n; i += 2) printf("-%c%c", s[i], s[i + 1]);
    puts("");
  } else {
    for (int i = 0; i < n; i += 2)
      printf("%s%c%c", i ? "-" : "", s[i], s[i + 1]);
    puts("");
  }
  return 0;
}

```