#include <bits/stdc++.h>
using namespace std;
char a[105];
int main() {
  int l;
  cin >> l;
  scanf("%s", a + 1);
  if (l % 2 == 0) {
    for (int i = 1; i <= l; i++) {
      cout << a[i];
      if (i % 2 == 0 && i != l) cout << "-";
    }
  } else {
    for (int i = 1; i <= l; i++) {
      cout << a[i];
      if (i % 2 == 0 && i != l - 1) cout << "-";
    }
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
char a[105];
int main() {
  int l;
  cin >> l;
  scanf("%s", a + 1);
  if (l % 2 == 0) {
    for (int i = 1; i <= l; i++) {
      cout << a[i];
      if (i % 2 == 0 && i != l) cout << "-";
    }
  } else {
    for (int i = 1; i <= l; i++) {
      cout << a[i];
      if (i % 2 == 0 && i != l - 1) cout << "-";
    }
  }
}

```