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:
//package main;

import java.io.*;
import java.util.*;
import java.math.*;

public final class Main {
    BufferedReader br;
    StringTokenizer stk;
    
    public static void main(String[] args) throws Exception {
        new Main().run();
    }
    
    {
        stk = null;
        br = new BufferedReader(new InputStreamReader(System.in));
    }
    
    long mod = 998244353;
    StringBuilder res = new StringBuilder(1000005);
    void run() throws Exception {
        String s = nt();
        int point = s.indexOf(".");
        if(point < 0) {
            point = s.length();
        }
        int first = -1;
        for(int i = 0; i < s.length(); i++) {
            if(s.charAt(i) != '0' && s.charAt(i) != '.') {
                first = i;
                break;
            }
        }
        
        if(first == -1) {
            System.out.println("0");
        } else {
            int count = 0;
            if(point < first) {
                count = first - point;
                count = -count;
            } else {
                count = point - first - 1;
            }
            res.append(s.charAt(first)).append(".");
            for(int i = first + 1; i < s.length(); i++) {
                if(s.charAt(i) == '.') {
                    continue;
                }
                res.append(s.charAt(i));
            }
            while(res.charAt(res.length() - 1) == '0') {
                res.deleteCharAt(res.length() - 1);
            }
            if(res.charAt(res.length() - 1) == '.') {
                res.deleteCharAt(res.length() - 1);
            }
            if(count != 0) {
                res.append("E").append(count);
            }
            System.out.println(res);
        }
    }
    
    //Reader & Writer
    String nt() throws Exception {
        if (stk == null || !stk.hasMoreTokens())
            stk = new StringTokenizer(br.readLine(), " ");
        return stk.nextToken();
    }

    char[] nc() throws Exception {
        return nt().toCharArray();
    }

    int ni() throws Exception {
        return Integer.parseInt(nt());
    }

    long nl() throws Exception {
        return Long.parseLong(nt());
    }
    
    double nd() throws Exception {
        return Double.parseDouble(nt());
    }
    
    //Some Misc methods
    int get(int l, int r, int[] a) {
        return l == 0 ? a[r] : a[r] - a[l - 1];
    }
    
    void shuffle(long[] a) {
        Random r = new Random();
        for(int i = 0; i < a.length; i++) {
            int idx = r.nextInt(a.length);
            long temp = a[i];
            a[i] = a[idx];
            a[idx] = temp;
        }
    }
    
    void reverse(long[] a) {
        for(int i = 0, j = a.length - 1; i < j; i++, j--) {
            long temp = a[i];
            a[i] = a[j];
            a[j] = temp;
        }
    }
}