#include <bits/stdc++.h>
int main() {
  char a[105];
  int i, n, c = 0;
  gets(a);
  n = strlen(a);
  for (i = 0; i < n - 1; i++) {
    if (a[i] == '/' && a[i + 1] == '/')
      continue;
    else {
      c++;
      printf("%c", a[i]);
    }
  }
  if (a[n - 1] != '/') printf("%c", a[n - 1]);
  if (!c) printf("/");
  printf("\n");
  return 0;
}

### Prompt

Please formulate 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>
int main() {
  char a[105];
  int i, n, c = 0;
  gets(a);
  n = strlen(a);
  for (i = 0; i < n - 1; i++) {
    if (a[i] == '/' && a[i + 1] == '/')
      continue;
    else {
      c++;
      printf("%c", a[i]);
    }
  }
  if (a[n - 1] != '/') printf("%c", a[n - 1]);
  if (!c) printf("/");
  printf("\n");
  return 0;
}

```