#include <bits/stdc++.h>
using namespace std;
void solve() {
  string s;
  cin >> s;
  int n = s.length();
  int i = 1;
  string ans = "/";
  while (i != n) {
    if (s[i] == '/' && s[i - 1] == '/') {
      i++;
    } else {
      ans += s[i];
      i++;
    }
  }
  n = ans.length();
  if (ans[n - 1] == '/' && n > 1) {
    ans = ans.substr(0, n - 1);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
void solve() {
  string s;
  cin >> s;
  int n = s.length();
  int i = 1;
  string ans = "/";
  while (i != n) {
    if (s[i] == '/' && s[i - 1] == '/') {
      i++;
    } else {
      ans += s[i];
      i++;
    }
  }
  n = ans.length();
  if (ans[n - 1] == '/' && n > 1) {
    ans = ans.substr(0, n - 1);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```