#include <bits/stdc++.h>
char a[110];
int main() {
  int n;
  while (scanf("%d", &n) > 0) {
    getchar();
    if (n % 2) {
      for (int i = 0; i < n - 3; i++) {
        scanf("%c", &a[i]);
        printf("%d", a[i] - '0');
        if (i % 2) printf("-");
      }
      scanf("%c%c%c", &a[0], &a[1], &a[2]);
      printf("%d%d%d\n", a[0] - '0', a[1] - '0', a[2] - '0');
    } else {
      for (int i = 0; i < n - 2; i++) {
        scanf("%c", &a[i]);
        printf("%d", a[i] - '0');
        if (i % 2) printf("-");
      }
      scanf("%c%c", &a[0], &a[1]);
      printf("%d%d\n", a[0] - '0', a[1] - '0');
    }
  }
  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>
char a[110];
int main() {
  int n;
  while (scanf("%d", &n) > 0) {
    getchar();
    if (n % 2) {
      for (int i = 0; i < n - 3; i++) {
        scanf("%c", &a[i]);
        printf("%d", a[i] - '0');
        if (i % 2) printf("-");
      }
      scanf("%c%c%c", &a[0], &a[1], &a[2]);
      printf("%d%d%d\n", a[0] - '0', a[1] - '0', a[2] - '0');
    } else {
      for (int i = 0; i < n - 2; i++) {
        scanf("%c", &a[i]);
        printf("%d", a[i] - '0');
        if (i % 2) printf("-");
      }
      scanf("%c%c", &a[0], &a[1]);
      printf("%d%d\n", a[0] - '0', a[1] - '0');
    }
  }
  return 0;
}

```