#include <bits/stdc++.h>
int main() {
  int i, r = 0;
  char s[110];
  scanf("%s", s);
  for (i = 0; s[i]; i++) {
    if (s[i] == '/') {
      while (s[i + 1] == '/') i++;
      if (s[i + 1]) {
        putchar('/');
        r = 1;
      }
    } else {
      putchar(s[i]);
      r = 1;
    }
  }
  if (r == 0) putchar('/');
  puts("");
}

### 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>
int main() {
  int i, r = 0;
  char s[110];
  scanf("%s", s);
  for (i = 0; s[i]; i++) {
    if (s[i] == '/') {
      while (s[i + 1] == '/') i++;
      if (s[i + 1]) {
        putchar('/');
        r = 1;
      }
    } else {
      putchar(s[i]);
      r = 1;
    }
  }
  if (r == 0) putchar('/');
  puts("");
}

```