#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<string> v;
  for (int i = 0; i < n; i += 2) {
    v.push_back(s.substr(i, 2));
    if (i == n - 1) {
      v[v.size() - 2] += v[v.size() - 1];
      v.pop_back();
    }
  }
  for (int i = 0; i < v.size(); i++) {
    cout << v[i];
    if (i != v.size() - 1) cout << "-";
  }
  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 main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<string> v;
  for (int i = 0; i < n; i += 2) {
    v.push_back(s.substr(i, 2));
    if (i == n - 1) {
      v[v.size() - 2] += v[v.size() - 1];
      v.pop_back();
    }
  }
  for (int i = 0; i < v.size(); i++) {
    cout << v[i];
    if (i != v.size() - 1) cout << "-";
  }
  return 0;
}

```