#include <bits/stdc++.h>
int main() {
  int n, l, i;
  char a[110];
  while (~scanf("%d%s", &n, a)) {
    int l = strlen(a);
    if (n == 1 || n == 2 || n == 3) {
      puts(a);
      continue;
    }
    if (n % 2 == 0) {
      for (int i = 0; i < n; ++i) {
        if (i % 2 == 0 && i != 0) {
          printf("-");
        }
        printf("%c", a[i]);
      }
      puts("");
    } else {
      printf("%c%c%c-", a[0], a[1], a[2]);
      int cnt = 0;
      for (int i = 3; i < n; ++i) {
        if (cnt % 2 == 0 && cnt != 0) {
          printf("-");
        }
        printf("%c", a[i]);
        ++cnt;
      }
      puts("");
    }
  }
}

### 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>
int main() {
  int n, l, i;
  char a[110];
  while (~scanf("%d%s", &n, a)) {
    int l = strlen(a);
    if (n == 1 || n == 2 || n == 3) {
      puts(a);
      continue;
    }
    if (n % 2 == 0) {
      for (int i = 0; i < n; ++i) {
        if (i % 2 == 0 && i != 0) {
          printf("-");
        }
        printf("%c", a[i]);
      }
      puts("");
    } else {
      printf("%c%c%c-", a[0], a[1], a[2]);
      int cnt = 0;
      for (int i = 3; i < n; ++i) {
        if (cnt % 2 == 0 && cnt != 0) {
          printf("-");
        }
        printf("%c", a[i]);
        ++cnt;
      }
      puts("");
    }
  }
}

```