#include <bits/stdc++.h>
using namespace std;
int main() {
  int n = 0;
  ios::sync_with_stdio(false);
  cin >> n;
  string s;
  cin >> s;
  if (n % 3 == 0) {
    for (int i = 0; i < n; ++i) {
      if (i && i % 3 == 0) {
        printf("-");
      }
      printf("%c", s[i]);
    }
    printf("\n");
  } else {
    int i = 0;
    for (; i < 2; ++i) {
      printf("%c", s[i]);
    }
    while (i + 2 < n) {
      printf("-%c%c%c", s[i], s[i + 1], s[i + 2]);
      i += 3;
    }
    if (i < n) {
      printf("-");
      for (; i < n; ++i) {
        printf("%c", s[i]);
      }
    }
    printf("\n");
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 n = 0;
  ios::sync_with_stdio(false);
  cin >> n;
  string s;
  cin >> s;
  if (n % 3 == 0) {
    for (int i = 0; i < n; ++i) {
      if (i && i % 3 == 0) {
        printf("-");
      }
      printf("%c", s[i]);
    }
    printf("\n");
  } else {
    int i = 0;
    for (; i < 2; ++i) {
      printf("%c", s[i]);
    }
    while (i + 2 < n) {
      printf("-%c%c%c", s[i], s[i + 1], s[i + 2]);
      i += 3;
    }
    if (i < n) {
      printf("-");
      for (; i < n; ++i) {
        printf("%c", s[i]);
      }
    }
    printf("\n");
  }
  return 0;
}

```