Algorithm


A. BerOS file system
time limit per test
2 seconds
memory limit per test
64 megabytes
input
standard input
output
standard output

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
Copy
//usr///local//nginx/sbin
output
Copy
/usr/local/nginx/sbin



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <iostream>

int main(){

    std::string path; getline(std::cin, path);
    std::string output; bool flag(0);
    for(int k = 0; k < path.size(); k++){
        if(path[k] != '/' || flag == 0){output += path[k];}
        if(path[k] == '/'){flag = 1;}
        else{flag = 0;}
    }

    if(flag && output.size() > 1){output = output.substr(0, output.size() - 1);}
    std::cout << output << std::endl;

    return 0;
}
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Input

x
+
cmd
//usr///local//nginx/sbin

Output

x
+
cmd
/usr/local/nginx/sbin
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Demonstration


Codeforces Solution-A. BerOS file system-Solution in C, C++, Java, Python

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