#include <bits/stdc++.h>
using namespace std;
long long int bigmod(long long int a, long long int p, long long int m) {
  if (p == 1) {
    return a % m;
  }
  if (p == 0) {
    return 1;
  }
  if (p % 2 == 0) {
    long long int ans = bigmod(a, p / 2, m);
    ans = (ans % m) * (ans % m);
    return ans % m;
  } else {
    long long int ans = bigmod(a, (p - 1), m);
    ans = (a % m) * (ans % m);
    return ans % m;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  while (cin >> s) {
    string ans;
    int p = s.size();
    for (int i = s.size() - 1; i >= 0; i--) {
      if (s[i] != '/') {
        break;
      } else {
        p = i;
      }
    }
    cout << s[0];
    for (int i = 1; i < p; i++) {
      if (s[i] == '/' && s[i - 1] == '/') {
        continue;
      } else {
        cout << s[i];
      }
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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 bigmod(long long int a, long long int p, long long int m) {
  if (p == 1) {
    return a % m;
  }
  if (p == 0) {
    return 1;
  }
  if (p % 2 == 0) {
    long long int ans = bigmod(a, p / 2, m);
    ans = (ans % m) * (ans % m);
    return ans % m;
  } else {
    long long int ans = bigmod(a, (p - 1), m);
    ans = (a % m) * (ans % m);
    return ans % m;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  while (cin >> s) {
    string ans;
    int p = s.size();
    for (int i = s.size() - 1; i >= 0; i--) {
      if (s[i] != '/') {
        break;
      } else {
        p = i;
      }
    }
    cout << s[0];
    for (int i = 1; i < p; i++) {
      if (s[i] == '/' && s[i - 1] == '/') {
        continue;
      } else {
        cout << s[i];
      }
    }
    cout << endl;
  }
  return 0;
}

```