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

### 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 n;
void del(int a[], int x) {
  for (int i = x; i < n - 1; i++) a[i] = a[i + 1];
  n--;
}
int main() {
  int n, x;
  char c;
  cin >> n;
  if (n % 2 == 0)
    x = 2;
  else
    x = 3;
  for (int i = 0; i < x; i++) {
    cin >> c;
    cout << c;
  }
  for (int i = 0; i < (n - x) / 2; i++) {
    cout << "-";
    cin >> c;
    cout << c;
    cin >> c;
    cout << c;
  }
}

```