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

### Prompt

Develop a solution in CPP to the problem described below:

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;
  v.clear();
  int i;
  for (i = 0; i < n; i += 2) {
    string S = s.substr(i, 2);
    v.push_back(S);
  }
  if (v.back().size() == 1) {
    v[v.size() - 2] += v.back();
    v.erase(v.begin() + v.size() - 1);
  }
  cout << v[0];
  for (i = 1; i < v.size(); i++) {
    cout << "-";
    cout << v[i];
  }
  cout << endl;
  return 0;
}

```