#include <bits/stdc++.h>
char s[105];
int main() {
  int n, f, t = 0;
  scanf("%d%s", &n, s);
  if (n % 2 == 0)
    f = 2;
  else
    f = 3;
  for (; t < f; ++t) printf("%c", s[t]);
  for (; s[t] != '\0'; ++t) {
    if ((t - f) % 2 == 0) printf("-");
    printf("%c", s[t]);
  }
  printf("\n");
  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>
char s[105];
int main() {
  int n, f, t = 0;
  scanf("%d%s", &n, s);
  if (n % 2 == 0)
    f = 2;
  else
    f = 3;
  for (; t < f; ++t) printf("%c", s[t]);
  for (; s[t] != '\0'; ++t) {
    if ((t - f) % 2 == 0) printf("-");
    printf("%c", s[t]);
  }
  printf("\n");
  return 0;
}

```