#include <bits/stdc++.h>
using namespace std;
vector<int> vr;
int main() {
  int n;
  string str;
  cin >> n >> str;
  if (n % 2 == 0) {
    for (int i = 0; i < str.length(); i++) {
      cout << str[i];
      if (i != 0 && i < str.length() - 1 && i % 2 != 0) cout << "-";
    }
    puts("");
  } else if (n != 3) {
    bool f = true;
    cout << str[0] << str[1] << "-";
    for (int i = 2; i < str.length(); i++) {
      cout << str[i];
      if (f && i == 4) {
        f = false;
        if (i < str.length() - 1) cout << "-";
        continue;
      }
      if (i != 0 && i < str.length() - 1 && i % 2 == 0 && !f) cout << "-";
    }
    puts("");
  } else
    cout << str << endl;
  return 0;
}

### Prompt

Generate 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;
vector<int> vr;
int main() {
  int n;
  string str;
  cin >> n >> str;
  if (n % 2 == 0) {
    for (int i = 0; i < str.length(); i++) {
      cout << str[i];
      if (i != 0 && i < str.length() - 1 && i % 2 != 0) cout << "-";
    }
    puts("");
  } else if (n != 3) {
    bool f = true;
    cout << str[0] << str[1] << "-";
    for (int i = 2; i < str.length(); i++) {
      cout << str[i];
      if (f && i == 4) {
        f = false;
        if (i < str.length() - 1) cout << "-";
        continue;
      }
      if (i != 0 && i < str.length() - 1 && i % 2 == 0 && !f) cout << "-";
    }
    puts("");
  } else
    cout << str << endl;
  return 0;
}

```