#include <bits/stdc++.h>
using namespace std;
int main() {
  long long i, j, l;
  char s[12345], ns[12345];
  scanf("%s", s);
  for (i = 0, j = 0; s[i]; i++) {
    if (s[i] != '/') {
      ns[j] = s[i];
      j++;
    } else {
      ns[j] = s[i];
      j++;
      for (i = i + 1; s[i] && s[i] == '/'; i++)
        ;
      i--;
    }
  }
  l = strlen(ns);
  if (ns[l - 1] == '/' && l != 1) ns[l - 1] = '\0';
  printf("%s\n", ns);
  return 0;
}

### Prompt

Generate 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() {
  long long i, j, l;
  char s[12345], ns[12345];
  scanf("%s", s);
  for (i = 0, j = 0; s[i]; i++) {
    if (s[i] != '/') {
      ns[j] = s[i];
      j++;
    } else {
      ns[j] = s[i];
      j++;
      for (i = i + 1; s[i] && s[i] == '/'; i++)
        ;
      i--;
    }
  }
  l = strlen(ns);
  if (ns[l - 1] == '/' && l != 1) ns[l - 1] = '\0';
  printf("%s\n", ns);
  return 0;
}

```