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.StringTokenizer;

/**
 * Created by peacefrog on 2/18/16.
 * 9:27 PM
 */
public class Task_B {
	final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
	PrintWriter out;
	long timeBegin, timeEnd;

	public void runIO() throws IOException {
		timeBegin = System.currentTimeMillis();

		InputStream inputStream;
		OutputStream outputStream;

		if (ONLINE_JUDGE) {
			inputStream = System.in;
			Reader.init(inputStream);
			outputStream = System.out;

			out = new PrintWriter(outputStream);
		} else {
			inputStream = new FileInputStream("/home/peacefrog/Dropbox/IdeaProjects/Problem Solving/input");
			Reader.init(inputStream);
			out = new PrintWriter(System.out);
		}
		solve();
		out.flush();
		out.close();
		timeEnd = System.currentTimeMillis();
		System.err.println("Time = " + (timeEnd - timeBegin));
	}

	/*
	 * Start Solution Here
	 */

	private void solve() throws IOException {
		double  a = Reader.nextDouble() , t = Reader.nextDouble();

		double factor = 1.000000011;

		out.println((a * Math.pow(factor, t)));

	}

	public static void main(String[] args) throws IOException {
		new Task_B().runIO();
	}

	static class Reader {
		static BufferedReader reader;
		static StringTokenizer tokenizer;

		/**
		 * call this method to initialize reader for InputStream
		 */
		static void init(InputStream input) {
			reader = new BufferedReader(new InputStreamReader(input));
			tokenizer = new StringTokenizer("");
		}

		/**
		 * get next word
		 */
		static String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		static String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return "";
		}

		static int nextChar() throws IOException {
			return reader.read();
		}

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

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

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

		static long[] nextLongArray(int n) throws IOException {
			long[] arr = new long[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nextLong();
			}

			return arr;
		}

		static int[] nextIntArray(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nextInt();
			}
			return arr;
		}
	}
}