Problem: 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

Solution (in JAVA):

Solution:
/*
 * Code Author: Jayesh Udhani
 * Institute  : DA-IICT 
 * 3rd year BTech ICT Student
 */

import java.io.*;
import java.math.*;
import java.util.*;
public class CFEDU14C
{
	static void solver(String s,PrintWriter out)
	{
		int i=0,n=s.length(),j;
		if(!s.contains("."))
		{
			while(i<n && s.charAt(i)=='0')
				i++;
			j=n-1;
			while(j>=0 && s.charAt(j)=='0')
				j--;
			String k=s.substring(i,j+1);
			if(k.substring(1).length()==0)
			{
				if(k.length()+n-j-2==0)
					out.println(k.charAt(0)+k.substring(1));
				else
					out.println(k.charAt(0)+k.substring(1)+"E"+(k.length()+n-j-2));
			}
			else
				out.println(k.charAt(0)+"."+k.substring(1)+"E"+(k.length()+n-j-2));
		}
		else
		{
			if(s.charAt(n-1)=='.')
				solver(s.substring(0, n-1),out);
			else
			{
				i=0;
				while(i<n && s.charAt(i)=='0')
					i++;
				j=n-1;
				while(j>=0 && s.charAt(j)=='0')
					j--;
				String k=s.substring(i, j+1);
				if(k.charAt(0)=='.')
				{
					int c=1;
					i=1;
					while(i<k.length() && k.charAt(i)=='0')
					{
						c++;
						i++;
					}
					String p=k.substring(c);
					if(p.substring(1).length()==0)
						out.println(p.charAt(0)+"E-"+c);
					else
						out.println(p.charAt(0)+"."+p.substring(1)+"E-"+c);
				}
				else if(k.charAt(k.length()-1)=='.')
					solver(k.substring(0, k.length()-1),out);
				else
				{
					String[] a=k.split("\\.");
					int dot=k.indexOf('.')-1;
					int l=k.length();
					if(dot==0)
						out.println(a[0].charAt(0)+"."+a[0].substring(1)+a[1]);
					else
						out.println(a[0].charAt(0)+"."+a[0].substring(1)+a[1]+"E"+(dot));
				}
			}
		}
	}
	public static void main(String args[])
	{
		InputReader in = new InputReader(System.in);
		OutputStream outputStream = System.out;
		PrintWriter out = new PrintWriter(outputStream);
		/*------------------------------My Code starts here------------------------------*/
		String s=in.next();
		solver(s,out);
		out.close();
		/*------------------------------The End------------------------------------------*/
	}
	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream inputstream) {
			reader = new BufferedReader(new InputStreamReader(inputstream));
			tokenizer = null;
		}

		public String nextLine(){
			String fullLine=null;
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					fullLine=reader.readLine();
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
				return fullLine;
			}
			return fullLine;
		}
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}