#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000;
const int N = 111123;
const double eps = 1e-10;
const double PI = acos(-1.0);
int n;
int ans;
char s[110];
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  int t = n / 2;
  int c = 1;
  int i;
  for (i = 1; i <= n && c < t; i += 2) {
    printf("%c%c-", s[i], s[i + 1]);
    c++;
  }
  for (; i <= n; i++) {
    printf("%c", s[i]);
  }
  cout << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
const int maxn = 100000;
const int N = 111123;
const double eps = 1e-10;
const double PI = acos(-1.0);
int n;
int ans;
char s[110];
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  int t = n / 2;
  int c = 1;
  int i;
  for (i = 1; i <= n && c < t; i += 2) {
    printf("%c%c-", s[i], s[i + 1]);
    c++;
  }
  for (; i <= n; i++) {
    printf("%c", s[i]);
  }
  cout << endl;
  return 0;
}

```