#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, j;
  cin >> n;
  char a[n + 1];
  for (int i = 1; i <= n; i++) cin >> a[i];
  int rem = n, x = 1;
  while (1) {
    if (rem > 5) {
      for (j = x; j <= x + 2; j++) cout << a[j];
      cout << "-";
      x = j;
      rem -= 3;
    } else if (rem == 5) {
      for (j = x; j <= x + 2; j++) cout << a[j];
      cout << "-";
      x = j;
      for (j = x; j <= n; j++) cout << a[j];
      break;
    } else if (rem == 4) {
      for (j = x; j <= x + 1; j++) cout << a[j];
      x = j;
      cout << "-";
      for (j = x; j <= n; j++) cout << a[j];
      break;
    } else if (rem == 3) {
      for (j = x; j <= n; j++) cout << a[j];
      break;
    } else {
      for (int i = 1; i <= n; i++) cout << a[i];
      break;
    }
  }
}

### 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, j;
  cin >> n;
  char a[n + 1];
  for (int i = 1; i <= n; i++) cin >> a[i];
  int rem = n, x = 1;
  while (1) {
    if (rem > 5) {
      for (j = x; j <= x + 2; j++) cout << a[j];
      cout << "-";
      x = j;
      rem -= 3;
    } else if (rem == 5) {
      for (j = x; j <= x + 2; j++) cout << a[j];
      cout << "-";
      x = j;
      for (j = x; j <= n; j++) cout << a[j];
      break;
    } else if (rem == 4) {
      for (j = x; j <= x + 1; j++) cout << a[j];
      x = j;
      cout << "-";
      for (j = x; j <= n; j++) cout << a[j];
      break;
    } else if (rem == 3) {
      for (j = x; j <= n; j++) cout << a[j];
      break;
    } else {
      for (int i = 1; i <= n; i++) cout << a[i];
      break;
    }
  }
}

```