#include <bits/stdc++.h>
using namespace std;
int main() {
  int jud0 = 0;
  char a;
  bool jud = false;
  while (scanf("%c", &a)) {
    if (a == '/' && !jud) {
      jud = true;
      jud0++;
    }
    if (a == '\n') {
      if (jud && jud0 == 1) cout << '/';
      cout << endl;
      return 0;
    }
    if (a != '/') {
      if (jud) cout << '/';
      cout << a;
      jud = false;
    }
  }
  cout << endl;
  return 0;
}

### 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>
using namespace std;
int main() {
  int jud0 = 0;
  char a;
  bool jud = false;
  while (scanf("%c", &a)) {
    if (a == '/' && !jud) {
      jud = true;
      jud0++;
    }
    if (a == '\n') {
      if (jud && jud0 == 1) cout << '/';
      cout << endl;
      return 0;
    }
    if (a != '/') {
      if (jud) cout << '/';
      cout << a;
      jud = false;
    }
  }
  cout << endl;
  return 0;
}

```