#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
  int n;
  scanf("%d", &n);
  string s, ans = "";
  cin >> s;
  if (n % 2 == 0) {
    for (int i = 0; i < n - 1; i = i + 2) {
      ans += s[i];
      ans += s[i + 1];
      ans += '-';
    }
    ans.erase(ans.length() - 1, 1);
  } else if (n % 3 == 0) {
    for (int i = 0; i < n - 2; i = i + 3) {
      ans += s[i];
      ans += s[i + 1];
      ans += s[i + 2];
      ans += '-';
    }
    ans.erase(ans.length() - 1, 1);
  } else {
    for (int i = 0; i < n - 4; i = i + 2) {
      ans += s[i];
      ans += s[i + 1];
      ans += '-';
    }
    ans += s[n - 3];
    ans += s[n - 2];
    ans += s[n - 1];
  }
  cout << ans << 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;
using ll = long long;
int main() {
  int n;
  scanf("%d", &n);
  string s, ans = "";
  cin >> s;
  if (n % 2 == 0) {
    for (int i = 0; i < n - 1; i = i + 2) {
      ans += s[i];
      ans += s[i + 1];
      ans += '-';
    }
    ans.erase(ans.length() - 1, 1);
  } else if (n % 3 == 0) {
    for (int i = 0; i < n - 2; i = i + 3) {
      ans += s[i];
      ans += s[i + 1];
      ans += s[i + 2];
      ans += '-';
    }
    ans.erase(ans.length() - 1, 1);
  } else {
    for (int i = 0; i < n - 4; i = i + 2) {
      ans += s[i];
      ans += s[i + 1];
      ans += '-';
    }
    ans += s[n - 3];
    ans += s[n - 2];
    ans += s[n - 1];
  }
  cout << ans << endl;
  return 0;
}

```