#include <bits/stdc++.h>
using namespace std;
int main() {
  int p = 1;
  char X = '/';
  string A;
  cin >> A;
  while (p) {
    for (int i = 0; i < A.size(); i++) {
      if (A[i] == X && A[i + 1] == X) {
        A.erase(A.begin() + i);
        p++;
      }
    }
    if (p > 1) {
      p = 1;
    } else
      p = 0;
  }
  int l = A.size();
  if ((l - 1) == 0 && A[0] == X) {
    cout << "/" << endl;
    return 0;
  }
  if (A[l - 1] == X) {
    A.erase(A.begin() + (l - 1));
  }
  cout << A << endl;
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

The new operating system BerOS has a nice feature. It is possible to use any number of characters '/' as a delimiter in path instead of one traditional '/'. For example, strings //usr///local//nginx/sbin// and /usr/local/nginx///sbin are equivalent. The character '/' (or some sequence of such characters) at the end of the path is required only in case of the path to the root directory, which can be represented as single character '/'.

A path called normalized if it contains the smallest possible number of characters '/'.

Your task is to transform a given path to the normalized form.

Input

The first line of the input contains only lowercase Latin letters and character '/' — the path to some directory. All paths start with at least one character '/'. The length of the given line is no more than 100 characters, it is not empty.

Output

The path in normalized form.

Examples

Input

//usr///local//nginx/sbin


Output

/usr/local/nginx/sbin

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int p = 1;
  char X = '/';
  string A;
  cin >> A;
  while (p) {
    for (int i = 0; i < A.size(); i++) {
      if (A[i] == X && A[i + 1] == X) {
        A.erase(A.begin() + i);
        p++;
      }
    }
    if (p > 1) {
      p = 1;
    } else
      p = 0;
  }
  int l = A.size();
  if ((l - 1) == 0 && A[0] == X) {
    cout << "/" << endl;
    return 0;
  }
  if (A[l - 1] == X) {
    A.erase(A.begin() + (l - 1));
  }
  cout << A << endl;
  return 0;
}

```