#include <bits/stdc++.h>
using namespace std;
int main() {
  char arr[101];
  int i;
  int Count = 0;
  cin >> arr;
  char arr1[101];
  int j = 0;
  for (i = 0; arr[i] != '\0'; i++) {
    if (arr[i] == '/') {
      if (Count == 0) {
        Count = 1;
        arr1[j++] = arr[i];
      }
    } else {
      Count = 0;
      arr1[j++] = arr[i];
    }
  }
  if (arr1[j - 1] == '/' && j > 1) j--;
  for (i = 0; i < j; i++) cout << arr1[i];
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 arr[101];
  int i;
  int Count = 0;
  cin >> arr;
  char arr1[101];
  int j = 0;
  for (i = 0; arr[i] != '\0'; i++) {
    if (arr[i] == '/') {
      if (Count == 0) {
        Count = 1;
        arr1[j++] = arr[i];
      }
    } else {
      Count = 0;
      arr1[j++] = arr[i];
    }
  }
  if (arr1[j - 1] == '/' && j > 1) j--;
  for (i = 0; i < j; i++) cout << arr1[i];
  return 0;
}

```