#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  string pd;
  cin >> pd;
  int v = 0;
  for (int i = 0; i < n; i++) {
    if (i == n - 2 && v == 1) {
      cout << pd[i] << pd[i + 1];
      break;
    }
    cout << pd[i];
    v++;
    if (v == 2 && i != n - 1) {
      cout << "-";
      v = 0;
    }
  }
  return 0;
}

### 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;
  cin >> n;
  string pd;
  cin >> pd;
  int v = 0;
  for (int i = 0; i < n; i++) {
    if (i == n - 2 && v == 1) {
      cout << pd[i] << pd[i + 1];
      break;
    }
    cout << pd[i];
    v++;
    if (v == 2 && i != n - 1) {
      cout << "-";
      v = 0;
    }
  }
  return 0;
}

```