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

### Prompt

Please provide a cpp coded solution to the problem described below:

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, j, m, n;
  char a[105];
  scanf("%s", a);
  n = strlen(a);
  printf("%c", a[0]);
  for (i = n - 1; i >= 0; i--) {
    if (a[i] == '/') n = n - 1;
    if (a[i] != '/') break;
  }
  for (i = 1; i < n; i++) {
    if (a[i - 1] != '/' || a[i] != '/') printf("%c", a[i]);
  }
  printf("\n");
  return 0;
}

```