#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  cin >> s;
  char a[100];
  int i, j;
  a[0] = s[0];
  j = 1;
  for (i = 1; i < s.length(); i++) {
    if (s[i] == '/') {
      if (a[j - 1] != '/') {
        a[j] = s[i];
        j++;
      }
    } else {
      a[j] = s[i];
      j++;
    }
  }
  if (a[j - 1] == '/' && (j - 1) != 0) j--;
  for (int k = 0; k < j; k++) cout << a[k];
}

### 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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  cin >> s;
  char a[100];
  int i, j;
  a[0] = s[0];
  j = 1;
  for (i = 1; i < s.length(); i++) {
    if (s[i] == '/') {
      if (a[j - 1] != '/') {
        a[j] = s[i];
        j++;
      }
    } else {
      a[j] = s[i];
      j++;
    }
  }
  if (a[j - 1] == '/' && (j - 1) != 0) j--;
  for (int k = 0; k < j; k++) cout << a[k];
}

```