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

```