#include <bits/stdc++.h>
using namespace std;
int n, a[10010];
char s[10010];
int main() {
  scanf("%d", &n);
  scanf("%s", s);
  for (int i = 0; i < n; i++) {
    a[i + 1] = s[i] - '0';
  }
  if (n % 2 == 0) {
    for (int i = 1; i <= n; i++) {
      printf("%d", a[i]);
      if (i % 2 == 0 && i != n) printf("-");
    }
  } else {
    printf("%d%d%d", a[1], a[2], a[3]);
    if (n == 3) return 0;
    printf("-");
    n -= 3;
    for (int i = 1; i <= n; i++) {
      printf("%d", a[i + 3]);
      if (i % 2 == 0 && i != n) printf("-");
    }
  }
  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>
using namespace std;
int n, a[10010];
char s[10010];
int main() {
  scanf("%d", &n);
  scanf("%s", s);
  for (int i = 0; i < n; i++) {
    a[i + 1] = s[i] - '0';
  }
  if (n % 2 == 0) {
    for (int i = 1; i <= n; i++) {
      printf("%d", a[i]);
      if (i % 2 == 0 && i != n) printf("-");
    }
  } else {
    printf("%d%d%d", a[1], a[2], a[3]);
    if (n == 3) return 0;
    printf("-");
    n -= 3;
    for (int i = 1; i <= n; i++) {
      printf("%d", a[i + 3]);
      if (i % 2 == 0 && i != n) printf("-");
    }
  }
  return 0;
}

```