#include <bits/stdc++.h>
using namespace std;
int n;
string a;
int main() {
  cin >> n;
  cin >> a;
  a = '0' + a;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n - (3 + n % 2); i += 2) cout << a[i] << a[i + 1] << "-";
  for (int i = n - (1 + n % 2); i <= n; i++) cout << a[i];
  return 0;
}

### 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>
using namespace std;
int n;
string a;
int main() {
  cin >> n;
  cin >> a;
  a = '0' + a;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n - (3 + n % 2); i += 2) cout << a[i] << a[i + 1] << "-";
  for (int i = n - (1 + n % 2); i <= n; i++) cout << a[i];
  return 0;
}

```