#include <bits/stdc++.h>
using namespace std;
char num[100], tmp;
int main() {
  int n;
  scanf("%u", &n);
  scanf("%c", &tmp);
  for (int i = 0; i < n; i++) scanf("%c", &num[i]);
  int a = 0, b = 0, off = 0;
  while (n % 2 != 0) b++, n -= 3;
  a = n / 2;
  for (int i = 0; a > 0 || b > 0; i++) {
    if (i > 0) printf("%c", '-');
    int len;
    if (i % 2 < 1)
      if (a > 0)
        a--, len = 2;
      else
        b--, len = 3;
    else if (b > 0)
      b--, len = 3;
    else
      a--, len = 2;
    for (int i = off; i < off + len; i++) printf("%c", num[i]);
    off += len;
  }
  printf("\n");
  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>
using namespace std;
char num[100], tmp;
int main() {
  int n;
  scanf("%u", &n);
  scanf("%c", &tmp);
  for (int i = 0; i < n; i++) scanf("%c", &num[i]);
  int a = 0, b = 0, off = 0;
  while (n % 2 != 0) b++, n -= 3;
  a = n / 2;
  for (int i = 0; a > 0 || b > 0; i++) {
    if (i > 0) printf("%c", '-');
    int len;
    if (i % 2 < 1)
      if (a > 0)
        a--, len = 2;
      else
        b--, len = 3;
    else if (b > 0)
      b--, len = 3;
    else
      a--, len = 2;
    for (int i = off; i < off + len; i++) printf("%c", num[i]);
    off += len;
  }
  printf("\n");
  return 0;
}

```