#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i = 0;
  string x, y;
  cin >> n;
  cin >> x;
  if (n % 2 != 0) {
    for (i = 0; i < 3; i++) y += x[i];
    y += '-';
  }
  for (; i < x.size(); i++) {
    y += x[i];
    i++;
    y += x[i];
    y += '-';
  }
  if (y[y.size() - 1] == '-') y.erase(y.size() - 1);
  cout << y << endl;
}

### 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, i = 0;
  string x, y;
  cin >> n;
  cin >> x;
  if (n % 2 != 0) {
    for (i = 0; i < 3; i++) y += x[i];
    y += '-';
  }
  for (; i < x.size(); i++) {
    y += x[i];
    i++;
    y += x[i];
    y += '-';
  }
  if (y[y.size() - 1] == '-') y.erase(y.size() - 1);
  cout << y << endl;
}

```