#include <bits/stdc++.h>
using namespace std;
int main() {
  int j, i, n;
  string s;
  while (cin >> s) {
    n = s.size();
    for (i = 0; i < n; i++) {
      while ((s[i] == '/' and s[i + 1] == '/')) {
        for (j = i + 1; j <= n; j++) s[j] = s[j + 1];
        n--;
      }
    }
    if (n > 1 and s[n - 1] == '/') n--;
    for (i = 0; i < n; i++) cout << s[i];
    cout << endl;
  }
  return 0;
}

### 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;
int main() {
  int j, i, n;
  string s;
  while (cin >> s) {
    n = s.size();
    for (i = 0; i < n; i++) {
      while ((s[i] == '/' and s[i + 1] == '/')) {
        for (j = i + 1; j <= n; j++) s[j] = s[j + 1];
        n--;
      }
    }
    if (n > 1 and s[n - 1] == '/') n--;
    for (i = 0; i < n; i++) cout << s[i];
    cout << endl;
  }
  return 0;
}

```