#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2147483647;
int i, j, k, l, n, m, t, x, y;
string sa, sb;
int main() {
  while (cin >> sa) {
    sb = "";
    i = 0;
    l = sa.length();
    j = l - 1;
    while (sa[j] == '/' && j > 0) {
      sa.erase(j, 1);
      j--;
    }
    l = sa.length();
    while (i < l) {
      if (sa[i] == '/') {
        sb += sa[i];
      }
      while (sa[i] == '/') {
        i++;
      }
      while (i < l && sa[i] != '/') {
        sb += sa[i++];
      }
    }
    cout << sb << endl;
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to 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;
const int MAXN = 2147483647;
int i, j, k, l, n, m, t, x, y;
string sa, sb;
int main() {
  while (cin >> sa) {
    sb = "";
    i = 0;
    l = sa.length();
    j = l - 1;
    while (sa[j] == '/' && j > 0) {
      sa.erase(j, 1);
      j--;
    }
    l = sa.length();
    while (i < l) {
      if (sa[i] == '/') {
        sb += sa[i];
      }
      while (sa[i] == '/') {
        i++;
      }
      while (i < l && sa[i] != '/') {
        sb += sa[i++];
      }
    }
    cout << sb << endl;
  }
  return 0;
}

```