#include <bits/stdc++.h>
using namespace std;
string s;
int main() {
  char pre = 'a';
  cin >> s;
  for (int i = 0; i < s.length(); ++i) {
    if (s[i] == '/' && pre == '/') {
      s[i] = '?';
      pre = '/';
      continue;
    }
    pre = s[i];
  }
  int j = s.length() - 1;
  while ((s[j] == '/' || s[j] == '?') && j >= 1) {
    s[j] = '?';
    --j;
  }
  for (int i = 0; i < s.length(); ++i) {
    if (s[i] != '?') cout << s[i];
  }
}

### Prompt

In Cpp, your task is to solve 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;
string s;
int main() {
  char pre = 'a';
  cin >> s;
  for (int i = 0; i < s.length(); ++i) {
    if (s[i] == '/' && pre == '/') {
      s[i] = '?';
      pre = '/';
      continue;
    }
    pre = s[i];
  }
  int j = s.length() - 1;
  while ((s[j] == '/' || s[j] == '?') && j >= 1) {
    s[j] = '?';
    --j;
  }
  for (int i = 0; i < s.length(); ++i) {
    if (s[i] != '?') cout << s[i];
  }
}

```