#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int a = 0;
  int ok = 0;
  if (n == 3 || n == 2) {
    string s;
    cin >> s;
    cout << s;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    a++;
    char x;
    cin >> x;
    cout << x;
    if (n % 2 != 0 && ok == 0) {
      if (a == 3) {
        a = 0;
        cout << "-";
        ok = 1;
      }
    }
    if (a == 2 && i != n && (ok || n % 2 == 0)) {
      cout << "-";
      a = 0;
    }
  }
  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;
int main() {
  int n;
  cin >> n;
  int a = 0;
  int ok = 0;
  if (n == 3 || n == 2) {
    string s;
    cin >> s;
    cout << s;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    a++;
    char x;
    cin >> x;
    cout << x;
    if (n % 2 != 0 && ok == 0) {
      if (a == 3) {
        a = 0;
        cout << "-";
        ok = 1;
      }
    }
    if (a == 2 && i != n && (ok || n % 2 == 0)) {
      cout << "-";
      a = 0;
    }
  }
  return 0;
}

```