#include <bits/stdc++.h>
const int MAXN = 100;
int N;
char chaine[MAXN + 1];
int main() {
  scanf("%d%s", &N, chaine);
  if (N % 2 == 0) {
    for (int i = 0; i < N - 2; i += 2)
      printf("%c%c-", chaine[i], chaine[i + 1]);
    printf("%c%c\n", chaine[N - 2], chaine[N - 1]);
  } else {
    for (int i = 0; i < N - 3; i += 2)
      printf("%c%c-", chaine[i], chaine[i + 1]);
    printf("%c%c%c\n", chaine[N - 3], chaine[N - 2], chaine[N - 1]);
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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>
const int MAXN = 100;
int N;
char chaine[MAXN + 1];
int main() {
  scanf("%d%s", &N, chaine);
  if (N % 2 == 0) {
    for (int i = 0; i < N - 2; i += 2)
      printf("%c%c-", chaine[i], chaine[i + 1]);
    printf("%c%c\n", chaine[N - 2], chaine[N - 1]);
  } else {
    for (int i = 0; i < N - 3; i += 2)
      printf("%c%c-", chaine[i], chaine[i + 1]);
    printf("%c%c%c\n", chaine[N - 3], chaine[N - 2], chaine[N - 1]);
  }
  return 0;
}

```