#include <bits/stdc++.h>
using namespace std;
int n;
char a[105];
bool flag;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  if (n % 2 == 0)
    n -= 2;
  else {
    n -= 3;
    flag = 1;
  }
  for (int i = 2; i <= n; i += 2) {
    cout << a[i - 1] << a[i] << "-";
  }
  if (flag == 0)
    cout << a[n + 1] << a[n + 2];
  else
    cout << a[n + 1] << a[n + 2] << a[n + 3];
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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;
char a[105];
bool flag;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  if (n % 2 == 0)
    n -= 2;
  else {
    n -= 3;
    flag = 1;
  }
  for (int i = 2; i <= n; i += 2) {
    cout << a[i - 1] << a[i] << "-";
  }
  if (flag == 0)
    cout << a[n + 1] << a[n + 2];
  else
    cout << a[n + 1] << a[n + 2] << a[n + 3];
  return 0;
}

```