#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  char c[100];
  bool flag = false;
  cin >> s;
  int j = 0, l = s.length();
  for (int i = 0; i < s.length(); ++i) {
    if (flag == false && s[i] == '/') {
      flag = true;
      c[j] = s[i];
      ++j;
    } else if (flag == true && s[i] == '/') {
      l--;
    } else {
      flag = false;
      c[j] = s[i];
      ++j;
    }
  }
  if (c[l - 1] == '/' && l > 1) --l;
  for (int i = 0; i < l; ++i) {
    cout << c[i];
  }
  cout << endl;
  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() {
  string s;
  char c[100];
  bool flag = false;
  cin >> s;
  int j = 0, l = s.length();
  for (int i = 0; i < s.length(); ++i) {
    if (flag == false && s[i] == '/') {
      flag = true;
      c[j] = s[i];
      ++j;
    } else if (flag == true && s[i] == '/') {
      l--;
    } else {
      flag = false;
      c[j] = s[i];
      ++j;
    }
  }
  if (c[l - 1] == '/' && l > 1) --l;
  for (int i = 0; i < l; ++i) {
    cout << c[i];
  }
  cout << endl;
  return 0;
}

```