#include <bits/stdc++.h>
using namespace std;
int main() {
  char dir[100 + 1], temp[100 + 1];
  int n, m = 0;
  cin.get(dir, 100 + 1, '\n');
  n = strlen(dir);
  for (int i = 0; i < n; ++i)
    if (dir[i] != '/' || m == 0 || temp[m - 1] != '/') temp[m++] = dir[i];
  if (m > 1 && temp[m - 1] == '/') m -= 1;
  temp[m] = '\0';
  cout << temp;
}

### Prompt

Generate 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() {
  char dir[100 + 1], temp[100 + 1];
  int n, m = 0;
  cin.get(dir, 100 + 1, '\n');
  n = strlen(dir);
  for (int i = 0; i < n; ++i)
    if (dir[i] != '/' || m == 0 || temp[m - 1] != '/') temp[m++] = dir[i];
  if (m > 1 && temp[m - 1] == '/') m -= 1;
  temp[m] = '\0';
  cout << temp;
}

```