#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  int n, i;
  string s;
  cin >> n;
  cin >> s;
  if (n == 2)
    cout << s;
  else if (n == 3)
    cout << s;
  else {
    if (s.length() % 2 == 0) {
      int x = s.length() / 2 - 1, cnt = 0;
      for (i = 2; i < s.length(); i += 3) {
        ++cnt;
        s.insert(i, "-");
        if (cnt == x) break;
      }
      cout << s << endl;
      return 0;
    } else {
      int x = s.length() / 2 - 1, cnt = 0;
      for (i = 2; i < s.length(); i += 3) {
        ++cnt;
        s.insert(i, "-");
        if (cnt == x) break;
      }
      cout << s << endl;
      return 0;
    }
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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() {
  ios_base::sync_with_stdio(0);
  int n, i;
  string s;
  cin >> n;
  cin >> s;
  if (n == 2)
    cout << s;
  else if (n == 3)
    cout << s;
  else {
    if (s.length() % 2 == 0) {
      int x = s.length() / 2 - 1, cnt = 0;
      for (i = 2; i < s.length(); i += 3) {
        ++cnt;
        s.insert(i, "-");
        if (cnt == x) break;
      }
      cout << s << endl;
      return 0;
    } else {
      int x = s.length() / 2 - 1, cnt = 0;
      for (i = 2; i < s.length(); i += 3) {
        ++cnt;
        s.insert(i, "-");
        if (cnt == x) break;
      }
      cout << s << endl;
      return 0;
    }
  }
  return 0;
}

```