#include <bits/stdc++.h>
using namespace std;
int main() {
  char a[10010];
  bool q = 0;
  int i, u, j;
  gets(a);
  cout << "/";
  u = strlen(a);
  j = u - 1;
  while (a[u - 1] == '/' && j >= 1) u--;
  for (i = 1; i < u; i++) {
    if (a[i] == '/' && a[i - 1] == '/') continue;
    cout << a[i];
  }
  cout << endl;
  return 0;
}

### Prompt

Your task is to create 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 a[10010];
  bool q = 0;
  int i, u, j;
  gets(a);
  cout << "/";
  u = strlen(a);
  j = u - 1;
  while (a[u - 1] == '/' && j >= 1) u--;
  for (i = 1; i < u; i++) {
    if (a[i] == '/' && a[i - 1] == '/') continue;
    cout << a[i];
  }
  cout << endl;
  return 0;
}

```