#include <bits/stdc++.h>
int main() {
  char s[101], t[101];
  int i, j = 1;
  scanf("%s", s);
  t[0] = s[0];
  for (i = 1; i < strlen(s);) {
    if (s[i - 1] == '/' && s[i - 1] == s[i])
      i++;
    else {
      t[j] = s[i];
      j++;
      i++;
    }
  }
  if (t[j - 1] == '/' && j != 1) j--;
  for (i = 0; i < j; i++) {
    printf("%c", t[i]);
  }
}

### Prompt

Develop a solution in Cpp 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() {
  char s[101], t[101];
  int i, j = 1;
  scanf("%s", s);
  t[0] = s[0];
  for (i = 1; i < strlen(s);) {
    if (s[i - 1] == '/' && s[i - 1] == s[i])
      i++;
    else {
      t[j] = s[i];
      j++;
      i++;
    }
  }
  if (t[j - 1] == '/' && j != 1) j--;
  for (i = 0; i < j; i++) {
    printf("%c", t[i]);
  }
}

```