#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i, j, l;
  char s[102];
  cin >> n;
  for (i = 0; i < n; ++i) cin >> s[i];
  if (n % 2 == 0) {
    for (i = 0; i < n; i++) {
      cout << s[i];
      if (i % 2 != 0 && i != n - 1) cout << "-";
    }
    return 0;
  }
  if (n % 3 == 0) {
    for (i = 0; i < n; i++) {
      cout << s[i];
      if ((i + 1) % 3 == 0 && i != n - 1) cout << "-";
    }
    return 0;
  }
  if (n % 3 == 2) {
    for (i = 0; i < n - 2; i++) {
      cout << s[i];
      if ((i + 1) % 3 == 0) cout << "-";
    }
    cout << s[n - 2] << s[n - 1];
    return 0;
  }
  if (n % 3 == 1) {
    for (i = 0; i < n - 4; i++) {
      cout << s[i];
      if ((i + 1) % 3 == 0) cout << "-";
    }
    cout << s[n - 4] << s[n - 3] << "-" << s[n - 2] << s[n - 1];
    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 n, i, j, l;
  char s[102];
  cin >> n;
  for (i = 0; i < n; ++i) cin >> s[i];
  if (n % 2 == 0) {
    for (i = 0; i < n; i++) {
      cout << s[i];
      if (i % 2 != 0 && i != n - 1) cout << "-";
    }
    return 0;
  }
  if (n % 3 == 0) {
    for (i = 0; i < n; i++) {
      cout << s[i];
      if ((i + 1) % 3 == 0 && i != n - 1) cout << "-";
    }
    return 0;
  }
  if (n % 3 == 2) {
    for (i = 0; i < n - 2; i++) {
      cout << s[i];
      if ((i + 1) % 3 == 0) cout << "-";
    }
    cout << s[n - 2] << s[n - 1];
    return 0;
  }
  if (n % 3 == 1) {
    for (i = 0; i < n - 4; i++) {
      cout << s[i];
      if ((i + 1) % 3 == 0) cout << "-";
    }
    cout << s[n - 4] << s[n - 3] << "-" << s[n - 2] << s[n - 1];
    return 0;
  }
}

```