#include <bits/stdc++.h>
using namespace std;
int n, c;
string s;
bool a;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  a = n % 2;
  cin >> s;
  for (int i = 0; i < n; i++) {
    cout << s[i];
    c++;
    if (i == n - 1) break;
    if (a) {
      if (c == 3) {
        cout << '-';
        c = 0;
        a = 0;
      }
    } else {
      if (c == 2) {
        cout << '-';
        c = 0;
      }
    }
  }
  return 0;
}

### Prompt

Please formulate 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, c;
string s;
bool a;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  a = n % 2;
  cin >> s;
  for (int i = 0; i < n; i++) {
    cout << s[i];
    c++;
    if (i == n - 1) break;
    if (a) {
      if (c == 3) {
        cout << '-';
        c = 0;
        a = 0;
      }
    } else {
      if (c == 2) {
        cout << '-';
        c = 0;
      }
    }
  }
  return 0;
}

```