#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  string error;
  getline(cin, error);
  string num;
  getline(cin, num);
  string ans;
  int it;
  if (n % 2 == 0) {
    it = 2;
  } else {
    it = 3;
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (cnt == 1) {
      it = 2;
    }
    cnt++;
    ans = num.substr(i, it);
    i = i + it - 1;
    if (i == n - 1) {
      cout << ans;
    } else {
      cout << ans << "-";
    }
  }
  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 error;
  getline(cin, error);
  string num;
  getline(cin, num);
  string ans;
  int it;
  if (n % 2 == 0) {
    it = 2;
  } else {
    it = 3;
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (cnt == 1) {
      it = 2;
    }
    cnt++;
    ans = num.substr(i, it);
    i = i + it - 1;
    if (i == n - 1) {
      cout << ans;
    } else {
      cout << ans << "-";
    }
  }
  return 0;
}

```