#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
char str[200];
char res[200];
int main() {
  gets(str);
  int len = strlen(str);
  int i = 0, j = 0;
  while (i < len) {
    if (str[i] == '/') {
      res[j++] = '/';
      while (str[i] == '/' && i < len) i++;
    } else
      res[j++] = str[i++];
  }
  res[j] = 0;
  if (res[j - 1] == '/' && j != 1) res[j - 1] = 0;
  printf("%s\n", res);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
char str[200];
char res[200];
int main() {
  gets(str);
  int len = strlen(str);
  int i = 0, j = 0;
  while (i < len) {
    if (str[i] == '/') {
      res[j++] = '/';
      while (str[i] == '/' && i < len) i++;
    } else
      res[j++] = str[i++];
  }
  res[j] = 0;
  if (res[j - 1] == '/' && j != 1) res[j - 1] = 0;
  printf("%s\n", res);
  return 0;
}

```