Problem: There are n students living in the campus. Every morning all students wake up at the same time and go to wash. There are m rooms with wash basins. The i-th of these rooms contains ai wash basins. Every student independently select one the rooms with equal probability and goes to it. After all students selected their rooms, students in each room divide into queues by the number of wash basins so that the size of the largest queue is the least possible. Calculate the expected value of the size of the largest queue among all rooms.

Input

The first line contains two positive integers n and m (1 ≤ n, m ≤ 50) — the amount of students and the amount of rooms. The second line contains m integers a1, a2, ... , am (1 ≤ ai ≤ 50). ai means the amount of wash basins in the i-th room.

Output

Output single number: the expected value of the size of the largest queue. Your answer must have an absolute or relative error less than 10 - 9.

Examples

Input

1 1
2


Output

1.00000000000000000000


Input

2 2
1 1


Output

1.50000000000000000000


Input

2 3
1 1 1


Output

1.33333333333333350000


Input

7 5
1 1 2 3 1


Output

2.50216960000000070000

Solution (in JAVA):

Solution:
import java.io.*;

import java.awt.geom.Point2D;
import java.text.*;
import java.math.*;
import java.util.*;

public class Main implements Runnable {

	final String filename = "";

	public void solve() throws Exception {
		int n = iread(), m = iread();

		double[][] C = new double[n + 1][n + 1];
		C[0][0] = 1.0;
		for (int i = 1; i <= n; i++) {
			C[i][0] = 1.0;
			C[i][i] = 1.0;
			for (int j = 1; j < i; j++)
				C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
		}

		double[][][] d = new double[m + 1][n + 1][n + 1];
		d[0][0][0] = 1.0;
		int[] a = new int[m];
		for (int i = 0; i < m; i++)
			a[i] = iread();
		for (int i = 0; i < m; i++) {
			for (int k = 0; k <= n; k++) {
				for (int j = 0; j <= n; j++) {
					if (d[i][j][k] == 0.0)
						continue;
					for (int j1 = 0; j + j1 <= n; j1++) {
						int k1 = Math.max(k, (j1 + a[i] - 1) / a[i]);
						d[i + 1][j + j1][k1] += d[i][j][k] * C[n - j][j1];
					}
				}
			}
		}
		double ans = 0.0;
		for (int j = 0; j <= n; j++) {
			ans += d[m][n][j] * j;
		}
		for (int j = 0; j < n; j++)
			ans /= m;
		DecimalFormat df = new DecimalFormat("0.0000000000");
		out.write(df.format(ans) + "\n");
	}

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new BufferedWriter(new OutputStreamWriter(System.out));
			// in = new BufferedReader(new FileReader(filename+".in"));
			// out = new BufferedWriter(new FileWriter(filename+".out"));
			solve();
			out.flush();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	public int iread() throws Exception {
		return Integer.parseInt(readword());
	}

	public double dread() throws Exception {
		return Double.parseDouble(readword());
	}

	public long lread() throws Exception {
		return Long.parseLong(readword());
	}

	BufferedReader in;

	BufferedWriter out;

	public String readword() throws IOException {
		StringBuilder b = new StringBuilder();
		int c;
		c = in.read();
		while (c >= 0 && c <= ' ')
			c = in.read();
		if (c < 0)
			return "";
		while (c > ' ') {
			b.append((char) c);
			c = in.read();
		}
		return b.toString();
	}

	public static void main(String[] args) {
		try {
			Locale.setDefault(Locale.US);
		} catch (Exception e) {

		}
		new Thread(new Main()).start();
		// new Thread(null, new Main(), "1", 1<<25).start();
	}
}