#include <bits/stdc++.h>
using namespace std;
int n, flag, a[107];
int main() {
  scanf("%d", &n);
  if (n % 2) flag = 1;
  for (int i = 1; i <= n; ++i) scanf("%1d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    printf("%d", a[i]);
    if (flag) {
      if (!(i % 2) && (i != n - 1)) putchar('-');
    } else if (!(i % 2) && (i != n))
      putchar('-');
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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, flag, a[107];
int main() {
  scanf("%d", &n);
  if (n % 2) flag = 1;
  for (int i = 1; i <= n; ++i) scanf("%1d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    printf("%d", a[i]);
    if (flag) {
      if (!(i % 2) && (i != n - 1)) putchar('-');
    } else if (!(i % 2) && (i != n))
      putchar('-');
  }
  return 0;
}

```