Codeforces 691C Exponential notation（模拟）

Exponential notation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a positive decimal number x.

Your task is to convert it to the "simple exponential notation".

Let x = a·10b, where 1 ≤ a < 10, then in general case the "simple exponential notation" looks like "aEb". If b equals to zero, the part "Eb" should be skipped. If a is an integer, it should be written without decimal point. Also there should not be extra zeroes in a and b.

Input

The only line contains the positive decimal number x. The length of the line will not exceed 106. Note that you are given too large number, so you can't use standard built-in data types "float", "double" and other.

Output

Print the only line — the "simple exponential notation" of the given number x.

Examples

input

16

output

1.6E1

input

01.23400

output

1.234

input

.100

output

1E-1

input

100.

output

1E2

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=1e6+5;
char s[maxn];
int main()
{
    while(cin>>s)
    {
        int len=strlen(s);
        int h=-1,t=len,p=len;//h第一个不为0的位置，t为最后一个不为0的位置，p是小数点的位置。
        for(int i=0;i<len;i++)
        {
            if(s[i]=='.')
                p=i;
            if(s[i]!='.'&&s[i]!='0'&&h==-1)
                h=i;
        }
        for(int i=len-1;i>=0;i--)
        {
            if(s[i]!='.'&&s[i]!='0')
            {
                t=i;
                break;
            }
        }
        if(h==-1)
            cout<<'0'<<endl;//说明都是0000.0000形式的
        else
        {
            cout<<s[h];//输出第一个数字
            if(h!=t)//后面存在有效数字
                cout<<'.';
            for(int i=h+1;i<=t;i++)
                if(s[i]!='.')
                    cout<<s[i];
            int e;//计算幂次
            if(h>p)
                e=p-h;
            else
                e=p-h-1;
            if(e)
                cout<<'E'<<e<<endl;
            else
                cout<<endl;
        }
    }
    return 0;
}