#include <bits/stdc++.h>
using namespace std;
char s[100000];
char r[100000];
int main() {
  int n;
  scanf("%d", &n);
  scanf("%s", s);
  char* p = s;
  while (strlen(p) > 3) {
    printf("%c%c", p[0], p[1]);
    p += 2;
    if (strlen(p)) printf("-");
  }
  if (strlen(p)) printf("%s", p);
  return 0;
}

### Prompt

Generate 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;
char s[100000];
char r[100000];
int main() {
  int n;
  scanf("%d", &n);
  scanf("%s", s);
  char* p = s;
  while (strlen(p) > 3) {
    printf("%c%c", p[0], p[1]);
    p += 2;
    if (strlen(p)) printf("-");
  }
  if (strlen(p)) printf("%s", p);
  return 0;
}

```