#include <bits/stdc++.h>
int main() {
  long long N, A[105], i;
  char S[105];
  scanf("%lld", &N);
  scanf("%s", &S);
  for (i = 0; i < N; i++) {
    A[i + 1] = S[i] - '0';
  }
  if (N % 3 == 0) {
    for (i = 1; i <= N;) {
      printf("%lld", A[i]);
      printf("%lld", A[i + 1]);
      printf("%lld", A[i + 2]);
      i = i + 3;
      if (i - 1 < N) printf("-");
    }
  } else if (N % 2 == 0) {
    for (i = 1; i <= N;) {
      printf("%lld", A[i]);
      printf("%lld", A[i + 1]);
      i = i + 2;
      if (i - 1 < N) printf("-");
    }
  } else {
    printf("%lld%lld%lld-", A[1], A[2], A[3]);
    for (i = 4; i <= N;) {
      printf("%lld", A[i]);
      printf("%lld", A[i + 1]);
      i = i + 2;
      if (i - 1 < N) printf("-");
    }
  }
  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>
int main() {
  long long N, A[105], i;
  char S[105];
  scanf("%lld", &N);
  scanf("%s", &S);
  for (i = 0; i < N; i++) {
    A[i + 1] = S[i] - '0';
  }
  if (N % 3 == 0) {
    for (i = 1; i <= N;) {
      printf("%lld", A[i]);
      printf("%lld", A[i + 1]);
      printf("%lld", A[i + 2]);
      i = i + 3;
      if (i - 1 < N) printf("-");
    }
  } else if (N % 2 == 0) {
    for (i = 1; i <= N;) {
      printf("%lld", A[i]);
      printf("%lld", A[i + 1]);
      i = i + 2;
      if (i - 1 < N) printf("-");
    }
  } else {
    printf("%lld%lld%lld-", A[1], A[2], A[3]);
    for (i = 4; i <= N;) {
      printf("%lld", A[i]);
      printf("%lld", A[i + 1]);
      i = i + 2;
      if (i - 1 < N) printf("-");
    }
  }
  return 0;
}

```