#include <bits/stdc++.h>
using namespace std;
int num[1000];
int main() {
  int n;
  scanf("%d", &n);
  string num1;
  cin >> num1;
  memset(num, 0, sizeof(num));
  for (int i = 0; i < n; i++) num[i] = num1[i] - '0';
  int flag = n % 2;
  if (flag == 0) {
    for (int i = 0; i < n; i = i + 2) {
      if (i != 0) {
        printf("-");
      }
      printf("%d%d", num[i], num[i + 1]);
    }
  } else {
    printf("%d%d%d", num[0], num[1], num[2]);
    for (int i = 3; i < n; i = i + 2) {
      printf("-");
      printf("%d%d", num[i], num[i + 1]);
    }
  }
  return 0;
}

### 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 num[1000];
int main() {
  int n;
  scanf("%d", &n);
  string num1;
  cin >> num1;
  memset(num, 0, sizeof(num));
  for (int i = 0; i < n; i++) num[i] = num1[i] - '0';
  int flag = n % 2;
  if (flag == 0) {
    for (int i = 0; i < n; i = i + 2) {
      if (i != 0) {
        printf("-");
      }
      printf("%d%d", num[i], num[i + 1]);
    }
  } else {
    printf("%d%d%d", num[0], num[1], num[2]);
    for (int i = 3; i < n; i = i + 2) {
      printf("-");
      printf("%d%d", num[i], num[i + 1]);
    }
  }
  return 0;
}

```