#include <bits/stdc++.h>
using namespace std;
long long int M = 1e9 + 7;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  long long int n = s.length();
  bool ok[n];
  memset(ok, true, sizeof(ok));
  long long int it = n - 1;
  while (s[it] == '/' && it >= 1) {
    ok[it] = 0;
    it--;
  }
  for (long long int i = 0; i < s.length(); ++i) {
    if (s[i] == '/') {
      long long int j = i + 1;
      while (s[j] == '/') {
        ok[j] = 0;
        j++;
      }
      i = j - 1;
    }
  }
  for (long long int i = 0; i < n; ++i) {
    if (ok[i]) cout << s[i];
  }
}

### 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;
long long int M = 1e9 + 7;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  long long int n = s.length();
  bool ok[n];
  memset(ok, true, sizeof(ok));
  long long int it = n - 1;
  while (s[it] == '/' && it >= 1) {
    ok[it] = 0;
    it--;
  }
  for (long long int i = 0; i < s.length(); ++i) {
    if (s[i] == '/') {
      long long int j = i + 1;
      while (s[j] == '/') {
        ok[j] = 0;
        j++;
      }
      i = j - 1;
    }
  }
  for (long long int i = 0; i < n; ++i) {
    if (ok[i]) cout << s[i];
  }
}

```