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

public class Main implements Runnable {
	private void solution() throws IOException {
		int n = in.nextInt();
		int m = in.nextInt();
		int[] a = new int[m];
		for (int i = 0; i < m; ++i) {
			a[i] = in.nextInt();
		}
		double res = 0;
		for (int i = 1; i <= n; ++i) {
			res += i * f(i, n, m, a);
		}
		out.printf(Locale.US, "%.12f\n", res);
	}

	private double f(int exact, int n, int m, int[] a) {
		Double[][][] am = new Double[n + 1][m + 1][2];
		double div = Math.pow(m, n);
		double res = solve(n, 0, a, exact, 0, am);
		return res / div;
	}

	private double solve(int n, int at, int[] a, int exact, int have, Double[][][] am) {
		if (at == a.length) {
			return (n == 0 && have == 1) ? 1 : 0;
		}
		if (am[n][at][have] != null) {
			return am[n][at][have];
		}
		double res = 0;
		for (int i = 0; i <= exact * a[at] && i <= n; ++i) {
			double cnt = CNK(n, i);
			int nhave = Math.max(have, (i + a[at] - 1) / a[at] == exact ? 1 : 0);
			res += cnt * solve(n - i, at + 1, a, exact, nhave, am);
		}
		return am[n][at][have] = res;
	}
	double[][] C = new double[100][100];
	{
		C[0][0] = 1;
		for (int i = 1; i < C.length; ++i) {
			C[i][0] = C[i][i] = 1;
			for (int j = 1; j < i; ++j) {
				C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
			}
		}
	}

	private double CNK(int n, int k) {
		return C[n][k];
	}

	public void run() {
		try {
			solution();
			in.reader.close();
			out.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	private class Scanner {
		BufferedReader reader;
		StringTokenizer tokenizer;

		public Scanner(Reader reader) {
			this.reader = new BufferedReader(reader);
			this.tokenizer = new StringTokenizer("");
		}

		public boolean hasNext() throws IOException {
			while (!tokenizer.hasMoreTokens()) {
				String next = reader.readLine();
				if (next == null) {
					return false;
				}
				tokenizer = new StringTokenizer(next);
			}
			return true;
		}

		public String next() throws IOException {
			hasNext();
			return tokenizer.nextToken();
		}

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

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

		public String nextLine() throws IOException {
			tokenizer = new StringTokenizer("");
			return reader.readLine();
		}
	}

	public static void main(String[] args) throws IOException {
		new Thread(null, new Main(), "", 1 << 28).start();
	}
	PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	Scanner in = new Scanner(new InputStreamReader(System.in));
}