#include <bits/stdc++.h>
char str[102];
char stt[102];
char sty[102];
int main() {
  int len, i, j, k = 0;
  scanf("%s", str);
  len = strlen(str);
  for (i = 0; i < len; i++) {
    if (str[i] == '/' && str[i + 1] == '/')
      continue;
    else
      stt[k++] = str[i];
  }
  if (k == 1)
    printf("%c\n", stt[0]);
  else if (stt[k - 1] == '/') {
    for (i = 0; i < k - 1; i++) printf("%c", stt[i]);
    printf("\n");
  } else
    printf("%s", stt);
  return 0;
}

### Prompt

Please create a solution in cpp 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>
char str[102];
char stt[102];
char sty[102];
int main() {
  int len, i, j, k = 0;
  scanf("%s", str);
  len = strlen(str);
  for (i = 0; i < len; i++) {
    if (str[i] == '/' && str[i + 1] == '/')
      continue;
    else
      stt[k++] = str[i];
  }
  if (k == 1)
    printf("%c\n", stt[0]);
  else if (stt[k - 1] == '/') {
    for (i = 0; i < k - 1; i++) printf("%c", stt[i]);
    printf("\n");
  } else
    printf("%s", stt);
  return 0;
}

```