#include <bits/stdc++.h>
using namespace std;
int n;
string txt, ans;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> txt;
  for (int i = 0; i < n - (2 + (n & 1)); i++) {
    if ((i & 1) == 0 && i) ans += '-';
    ans += txt[i];
  }
  if (n > 3) ans += '-';
  for (int i = n - (2 + (n & 1)); i < n; i++) ans += txt[i];
  return cout << ans << endl, 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 n;
string txt, ans;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> txt;
  for (int i = 0; i < n - (2 + (n & 1)); i++) {
    if ((i & 1) == 0 && i) ans += '-';
    ans += txt[i];
  }
  if (n > 3) ans += '-';
  for (int i = n - (2 + (n & 1)); i < n; i++) ans += txt[i];
  return cout << ans << endl, 0;
}

```