#include <bits/stdc++.h>
using namespace std;
int main() {
  int i, j, k, m, ln, ans, sm, cnt, ar[10000];
  string s;
  while (cin >> s) {
    k = 0;
    ln = s.size();
    string s1 = "";
    for (i = 0; s[i]; i++) {
      if (k == 0 && s[i] == '/') {
        s1 += "/";
        k = 1;
      }
      if (isalpha(s[i])) {
        s1 += s[i];
        k = 0;
      }
    }
    if (s1.size() > 1 && s1[s1.size() - 1] == '/') s1.erase(s1.end() - 1);
    cout << s1;
    cout << endl;
  }
}

### Prompt

Please formulate 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;
int main() {
  int i, j, k, m, ln, ans, sm, cnt, ar[10000];
  string s;
  while (cin >> s) {
    k = 0;
    ln = s.size();
    string s1 = "";
    for (i = 0; s[i]; i++) {
      if (k == 0 && s[i] == '/') {
        s1 += "/";
        k = 1;
      }
      if (isalpha(s[i])) {
        s1 += s[i];
        k = 0;
      }
    }
    if (s1.size() > 1 && s1[s1.size() - 1] == '/') s1.erase(s1.end() - 1);
    cout << s1;
    cout << endl;
  }
}

```