#include <bits/stdc++.h>
using namespace std;
int N, M, T;
int main() {
  int i, j, k;
  char ch;
  bool ok = true;
  char s[127];
  scanf("%s", s);
  k = strlen(s);
  for (i = k - 1; i > 0 && s[i] == '/'; i--) k = i;
  for (i = 0; i < k; i++) {
    ch = s[i];
    if (ch == '/') {
      if (ok) printf("%c", ch);
      ok = false;
    } else {
      printf("%c", ch);
      ok = true;
    }
  }
}

### 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;
int N, M, T;
int main() {
  int i, j, k;
  char ch;
  bool ok = true;
  char s[127];
  scanf("%s", s);
  k = strlen(s);
  for (i = k - 1; i > 0 && s[i] == '/'; i--) k = i;
  for (i = 0; i < k; i++) {
    ch = s[i];
    if (ch == '/') {
      if (ok) printf("%c", ch);
      ok = false;
    } else {
      printf("%c", ch);
      ok = true;
    }
  }
}

```