Problem: The city administration of IT City decided to fix up a symbol of scientific and technical progress in the city's main square, namely an indicator board that shows the effect of Moore's law in real time.

Moore's law is the observation that the number of transistors in a dense integrated circuit doubles approximately every 24 months. The implication of Moore's law is that computer performance as function of time increases exponentially as well.

You are to prepare information that will change every second to display on the indicator board. Let's assume that every second the number of transistors increases exactly 1.000000011 times.

Input

The only line of the input contains a pair of integers n (1000 ≤ n ≤ 10 000) and t (0 ≤ t ≤ 2 000 000 000) — the number of transistors in the initial time and the number of seconds passed since the initial time.

Output

Output one number — the estimate of the number of transistors in a dence integrated circuit in t seconds since the initial time. The relative error of your answer should not be greater than 10 - 6.

Examples

Input

1000 1000000


Output

1011.060722383550382782399454922040

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class b
{
	public static double fast_pow(double n, int t) {
		if(t == 0)
			return 1;
		double ans = 1, exp = n;
		while(t > 0) {
			if((t & 1) == 1) {
				ans *= exp;
			}
			exp *= exp;
			t >>= 1;
		}
		
		return ans;
	}
	
    public static void main(String[] args)
    {
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
        
        double n = sc.nextInt();
        int t = sc.nextInt();
        
        //out.println(fast_pow(2, 7));
        
        out.printf("%.15f\n", n * fast_pow(1.000000011, t));
       

        // Start writing your solution here.

        /*
        int n      = sc.nextInt();        // read input as integer
        long k     = sc.nextLong();       // read input as long
        double d   = sc.nextDouble();     // read input as double
        String str = sc.next();           // read input as String
        String s   = sc.nextLine();       // read whole line as String

    	while(sc.hasNext()) {
    		...
    	}

        int result = 3*n;
        out.println(result);                    // print via PrintWriter
        */

        // Stop writing your solution here.
        out.close();
    }

    // PrintWriter for faster output
    public static PrintWriter out;

    // MyScanner class for faster input
    public static class MyScanner
    {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner()
        {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        boolean hasNext()
        {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                    return false;
                }
            }
            return true;
        }

        String next()
        {
            if (hasNext())
                return st.nextToken();
            return null;
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}