#include <bits/stdc++.h>
using namespace std;
int main() {
  char str[105];
  scanf("%s", str);
  int len = strlen(str);
  int flag = 1, in = 0;
  for (int i = 0; i < len; i++) {
    if (str[i] != '/') {
      if (flag == 1) {
        printf("/");
        flag = 2;
        in = 1;
      }
      printf("%c", str[i]);
    } else
      flag = 1;
  }
  if (!in) printf("/");
  return 0;
}

### Prompt

Your challenge is to write 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() {
  char str[105];
  scanf("%s", str);
  int len = strlen(str);
  int flag = 1, in = 0;
  for (int i = 0; i < len; i++) {
    if (str[i] != '/') {
      if (flag == 1) {
        printf("/");
        flag = 2;
        in = 1;
      }
      printf("%c", str[i]);
    } else
      flag = 1;
  }
  if (!in) printf("/");
  return 0;
}

```