#include <bits/stdc++.h>
using namespace std;
string s;
bool read() {
  if (!(cin >> s)) return false;
  return true;
}
bool cmp(char &a, char &b) { return a == b && a == '/'; }
void solve() {
  s.resize(unique(s.begin(), s.end(), cmp) - s.begin());
  if (int(s.size()) > 1) {
    while (s[int(s.size()) - 1] == '/') s.resize(int(s.size()) - 1);
  }
  puts(s.c_str());
}
int main() {
  assert(read());
  solve();
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
string s;
bool read() {
  if (!(cin >> s)) return false;
  return true;
}
bool cmp(char &a, char &b) { return a == b && a == '/'; }
void solve() {
  s.resize(unique(s.begin(), s.end(), cmp) - s.begin());
  if (int(s.size()) > 1) {
    while (s[int(s.size()) - 1] == '/') s.resize(int(s.size()) - 1);
  }
  puts(s.c_str());
}
int main() {
  assert(read());
  solve();
  return 0;
}

```