#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lb = 1000;
void vrikodara(long long n = 12) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(n) << fixed;
}
void solve() {
  string s;
  cin >> s;
  string ans = "";
  long long sz = s.length();
  long long idx = 0;
  bool f = false;
  while (idx < sz) {
    if (s[idx] != '/') {
      f = true;
      ans += s[idx];
      idx++;
      continue;
    }
    while (idx < sz && s[idx] == '/') idx++;
    if (idx == sz && f) continue;
    ans += '/';
  }
  cout << ans << '\n';
}
int32_t main() {
  vrikodara();
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lb = 1000;
void vrikodara(long long n = 12) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(n) << fixed;
}
void solve() {
  string s;
  cin >> s;
  string ans = "";
  long long sz = s.length();
  long long idx = 0;
  bool f = false;
  while (idx < sz) {
    if (s[idx] != '/') {
      f = true;
      ans += s[idx];
      idx++;
      continue;
    }
    while (idx < sz && s[idx] == '/') idx++;
    if (idx == sz && f) continue;
    ans += '/';
  }
  cout << ans << '\n';
}
int32_t main() {
  vrikodara();
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```