#include <bits/stdc++.h>
using namespace std;
int n;
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  cin >> s;
  int byk;
  if (n & 1) {
    byk = (n - 3) / 2;
  } else
    byk = n / 2;
  for (int i = 0; i < byk * 2; i++) {
    if (i != 0 && i % 2 == 0) cout << '-';
    cout << s[i];
  }
  if (n & 1) {
    if (n > 3) cout << '-';
    cout << s[n - 3] << s[n - 2] << s[n - 1];
  }
  cout << endl;
}

### Prompt

Please create a solution in Cpp 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 s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  cin >> s;
  int byk;
  if (n & 1) {
    byk = (n - 3) / 2;
  } else
    byk = n / 2;
  for (int i = 0; i < byk * 2; i++) {
    if (i != 0 && i % 2 == 0) cout << '-';
    cout << s[i];
  }
  if (n & 1) {
    if (n > 3) cout << '-';
    cout << s[n - 3] << s[n - 2] << s[n - 1];
  }
  cout << endl;
}

```