#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, i, j, k, l = -1;
  string s;
  cin >> n >> s;
  k = n / 2, k--;
  for (i = 1; i <= k; i++) printf("%c%c-", s[l + 1], s[l + 2]), l += 2;
  cout << s[l + 1] << s[2 + l];
  if (n % 2) cout << s[3 + l] << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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, m, i, j, k, l = -1;
  string s;
  cin >> n >> s;
  k = n / 2, k--;
  for (i = 1; i <= k; i++) printf("%c%c-", s[l + 1], s[l + 2]), l += 2;
  cout << s[l + 1] << s[2 + l];
  if (n % 2) cout << s[3 + l] << endl;
  return 0;
}

```