#include <bits/stdc++.h>
using namespace std;
int main() {
  int ks = 1, tt, i, j, k, n;
  char a[104];
  while (scanf("%d", &n) == 1) {
    scanf("%s", a);
    k = n / 2;
    tt = k * 2;
    for (i = 0; i < tt; i += 2) {
      if (i) printf("-");
      printf("%c%c", a[i], a[i + 1]);
    }
    if (n & 1) cout << a[n - 1];
    cout << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
int main() {
  int ks = 1, tt, i, j, k, n;
  char a[104];
  while (scanf("%d", &n) == 1) {
    scanf("%s", a);
    k = n / 2;
    tt = k * 2;
    for (i = 0; i < tt; i += 2) {
      if (i) printf("-");
      printf("%c%c", a[i], a[i + 1]);
    }
    if (n & 1) cout << a[n - 1];
    cout << endl;
  }
  return 0;
}

```