#include <bits/stdc++.h>
using namespace std;
int main() {
  char ph[100];
  int i, j, k, t;
  while (cin >> t) {
    if (t % 2 == 0)
      for (i = 1; i <= t; i++) {
        cin >> ph[i];
        cout << ph[i];
        if (i % 2 == 0 && i < t) cout << "-";
        if (i == t) cout << endl;
      }
    if (t % 2 != 0) {
      for (j = 1; j <= t / 2; j++) {
        cin >> ph[j];
        cout << ph[j];
        if (j % 2 == 0 && j < t / 2) cout << "-";
      }
      for (k = t / 2 + 1; k <= t; k++) {
        cin >> ph[j];
        cout << ph[j];
        if ((k + 1) % 2 == 0 && k < t) cout << "-";
        if (k == t) cout << endl;
      }
    }
  }
}

### 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 main() {
  char ph[100];
  int i, j, k, t;
  while (cin >> t) {
    if (t % 2 == 0)
      for (i = 1; i <= t; i++) {
        cin >> ph[i];
        cout << ph[i];
        if (i % 2 == 0 && i < t) cout << "-";
        if (i == t) cout << endl;
      }
    if (t % 2 != 0) {
      for (j = 1; j <= t / 2; j++) {
        cin >> ph[j];
        cout << ph[j];
        if (j % 2 == 0 && j < t / 2) cout << "-";
      }
      for (k = t / 2 + 1; k <= t; k++) {
        cin >> ph[j];
        cout << ph[j];
        if ((k + 1) % 2 == 0 && k < t) cout << "-";
        if (k == t) cout << endl;
      }
    }
  }
}

```