#include <bits/stdc++.h>
using namespace std;
long long int _xor(long long int a, long long int b) { return a ^ b; }
long long int _and(long long int a, long long int b) { return a & b; }
long long int _or(long long int a, long long int b) { return a | b; }
long long int _not(long long int a) { return ~a; }
long long int ai[1000], pi[2005];
int main() {
  long long int b, c = 0, x, y, z, t, i, j, k, l1, l2, m, n, c1 = 0, c2 = 0, q,
                   f = 1;
  scanf("%lld", &n);
  string str;
  cin >> str;
  if (n % 2) {
    cout << str[0] << str[1] << str[2];
    if (n > 3) {
      for (i = 3; i <= n - 4; i += 2) cout << '-' << str[i] << str[i + 1];
      cout << '-' << str[n - 2] << str[n - 1];
    }
  } else {
    for (i = 0; i <= n - 4; i += 2) cout << str[i] << str[i + 1] << '-';
    cout << str[n - 2] << str[n - 1];
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
long long int _xor(long long int a, long long int b) { return a ^ b; }
long long int _and(long long int a, long long int b) { return a & b; }
long long int _or(long long int a, long long int b) { return a | b; }
long long int _not(long long int a) { return ~a; }
long long int ai[1000], pi[2005];
int main() {
  long long int b, c = 0, x, y, z, t, i, j, k, l1, l2, m, n, c1 = 0, c2 = 0, q,
                   f = 1;
  scanf("%lld", &n);
  string str;
  cin >> str;
  if (n % 2) {
    cout << str[0] << str[1] << str[2];
    if (n > 3) {
      for (i = 3; i <= n - 4; i += 2) cout << '-' << str[i] << str[i + 1];
      cout << '-' << str[n - 2] << str[n - 1];
    }
  } else {
    for (i = 0; i <= n - 4; i += 2) cout << str[i] << str[i + 1] << '-';
    cout << str[n - 2] << str[n - 1];
  }
  return 0;
}

```