#include <bits/stdc++.h>
using namespace std;
int main() {
  char str[1000];
  while (scanf("%s", str) != EOF) {
    int len = strlen(str);
    for (int i = 0; i < len; i++) {
      if (str[i] == '/') {
        i++;
        while (i < len && str[i] == '/') {
          str[i] = '.';
          i++;
        }
        i--;
      }
    }
    int k = 0;
    char ts[1000];
    for (int i = 0; i < len; i++)
      if (str[i] != '.') ts[k++] = str[i];
    while (k > 0 && ts[k - 1] == '/') k--;
    ts[k] = '\0';
    if (k == 0)
      printf("/\n");
    else
      printf("%s\n", ts);
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 main() {
  char str[1000];
  while (scanf("%s", str) != EOF) {
    int len = strlen(str);
    for (int i = 0; i < len; i++) {
      if (str[i] == '/') {
        i++;
        while (i < len && str[i] == '/') {
          str[i] = '.';
          i++;
        }
        i--;
      }
    }
    int k = 0;
    char ts[1000];
    for (int i = 0; i < len; i++)
      if (str[i] != '.') ts[k++] = str[i];
    while (k > 0 && ts[k - 1] == '/') k--;
    ts[k] = '\0';
    if (k == 0)
      printf("/\n");
    else
      printf("%s\n", ts);
  }
  return 0;
}

```