BerOS file system CodeForces - 20A

BerOS file system

CodeForces - 20A

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.

Example

Input

//usr///local//nginx/sbin

Output

/usr/local/nginx/sbin

给你一个 字符数不超过100的 仅由小写字母与字符'/'构成的 字符串

然后把字符串的一些不必要的字符去掉了

  1.如果出现连续的'/'那么只需要一个'/'即可 

  2.把路径名最后面那些'/'去掉了，除非路径名仅由一个'/'组成

code:

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
char str[1100];
int main(){
    while(~scanf("%s",str)){
        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 ans[1000];
        for(int i = 0; i < len; i++){
            if(str[i] != '.'){
                ans[k++] = str[i];
            }
        }
        while(k > 0 && ans[k-1] == '/') k--;
        ans[k] = 0;
        if(k == 0) printf("/\n");
        else printf("%s\n",ans);
    }
    return 0;
}