#include <bits/stdc++.h>
using namespace std;
int N;
char S[111];
int main() {
  cin >> N >> S;
  for (int(a) = (0); (a) < (N); (a)++) {
    for (int(b) = (0); (b) < (N); (b)++) {
      if (a * 2 + b * 3 == N) {
        int foi = 0, pos = 0;
        for (int(aa) = (0); (aa) < (a); (aa)++) {
          if (foi++) {
            printf("-");
          }
          for (int(k) = (0); (k) < (2); (k)++) {
            printf("%c", S[pos++]);
          }
        }
        for (int(bb) = (0); (bb) < (b); (bb)++) {
          if (foi++) {
            printf("-");
          }
          for (int(k) = (0); (k) < (3); (k)++) {
            printf("%c", S[pos++]);
          }
        }
        printf("\n");
        return 0;
      }
    }
  }
  return (0);
}

### 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 N;
char S[111];
int main() {
  cin >> N >> S;
  for (int(a) = (0); (a) < (N); (a)++) {
    for (int(b) = (0); (b) < (N); (b)++) {
      if (a * 2 + b * 3 == N) {
        int foi = 0, pos = 0;
        for (int(aa) = (0); (aa) < (a); (aa)++) {
          if (foi++) {
            printf("-");
          }
          for (int(k) = (0); (k) < (2); (k)++) {
            printf("%c", S[pos++]);
          }
        }
        for (int(bb) = (0); (bb) < (b); (bb)++) {
          if (foi++) {
            printf("-");
          }
          for (int(k) = (0); (k) < (3); (k)++) {
            printf("%c", S[pos++]);
          }
        }
        printf("\n");
        return 0;
      }
    }
  }
  return (0);
}

```