#include <bits/stdc++.h>
using namespace std;
int main(int argc, char* argv[]) {
  char str[105];
  cin.getline(str, 104);
  for (unsigned i = 1; i < strlen(str) + 1; i++) {
    if ((str[i] == str[i - 1]) && (str[i - 1] == '/')) {
      for (unsigned j = i; j < strlen(str); j++) {
        str[j] = str[j + 1];
      }
      i--;
    }
  }
  if ((str[strlen(str) - 1] == '/') && (strlen(str) >= 2)) {
    str[strlen(str) - 1] = 0;
  }
  cout << str << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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(int argc, char* argv[]) {
  char str[105];
  cin.getline(str, 104);
  for (unsigned i = 1; i < strlen(str) + 1; i++) {
    if ((str[i] == str[i - 1]) && (str[i - 1] == '/')) {
      for (unsigned j = i; j < strlen(str); j++) {
        str[j] = str[j + 1];
      }
      i--;
    }
  }
  if ((str[strlen(str) - 1] == '/') && (strlen(str) >= 2)) {
    str[strlen(str) - 1] = 0;
  }
  cout << str << endl;
  return 0;
}

```