#include <bits/stdc++.h>
int main() {
  char input[107];
  char output[107];
  int i, o;
  while (1 == scanf("%s", input)) {
    i = 0;
    o = 0;
    for (; input[i]; i++) {
      if (input[i] == '/') {
        output[o++] = input[i];
        while (input[++i] == '/')
          ;
        --i;
      } else
        output[o++] = input[i];
    }
    while (o > 0 && output[--o] == '/')
      ;
    output[o + 1] = 0;
    printf("%s", output);
    printf("\n");
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
int main() {
  char input[107];
  char output[107];
  int i, o;
  while (1 == scanf("%s", input)) {
    i = 0;
    o = 0;
    for (; input[i]; i++) {
      if (input[i] == '/') {
        output[o++] = input[i];
        while (input[++i] == '/')
          ;
        --i;
      } else
        output[o++] = input[i];
    }
    while (o > 0 && output[--o] == '/')
      ;
    output[o + 1] = 0;
    printf("%s", output);
    printf("\n");
  }
  return 0;
}

```