Code Forces 20A BerOS file system

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

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

output

/usr/local/nginx/sbin

遇到多个////

只输出一个/

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
char a[105];
char b[105];
int main()
{
    gets(a);
    bool tag=0;
    int len=strlen(a);
    int cnt=0;
    for(int i=0;i<len;i++)
    {
        if(a[i]!='/')
        {
           // cout<<a[i];
            b[cnt++]=a[i];
            tag=0;
        }

        else
        {
            if(!tag)
            {
                //cout<<a[i];
                b[cnt++]=a[i];
                tag=1;
            }
        }
    }
    if(b[0]!='/')
        cout<<'/';
    for(int i=0;i<cnt;i++)
    {
        if(i==cnt-1&&b[i]=='/'&&cnt!=1)
            continue;
       cout<<b[i];
    }
    cout<<endl;
    return 0;
}