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.util.*;

public class C implements Runnable {
	public static void main(String[] args) {
		new C().run();
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		boolean eof;
		String buf;

		public FastScanner(String fileName) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(fileName));
			nextToken();
		}

		public FastScanner(InputStream stream) {
			br = new BufferedReader(new InputStreamReader(stream));
			nextToken();
		}

		String nextToken() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					eof = true;
					break;
				}
			}
			String ret = buf;
			buf = eof ? "-1" : st.nextToken();
			return ret;
		}

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

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

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

		void close() {
			try {
				br.close();
			} catch (Exception e) {

			}
		}

		boolean isEOF() {
			return eof;
		}
	}

	FastScanner sc;
	PrintWriter out;

	public void run() {
		Locale.setDefault(Locale.US);
		try {
			sc = new FastScanner(System.in);
			out = new PrintWriter(System.out);
			solve();
			sc.close();
			out.close();
		} catch (Throwable e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	int nextInt() {
		return sc.nextInt();
	}

	String nextToken() {
		return sc.nextToken();
	}

	long nextLong() {
		return sc.nextLong();
	}

	double nextDouble() {
		return sc.nextDouble();
	}

	void solve() {
		int n = nextInt();
		int m = nextInt();
		int[] a = new int[m];
		for (int i = 0; i < a.length; i++) {
			a[i] = nextInt();
		}
		double[][][] dp = new double[m + 1][n + 1][n + 1];
		for (int i = 0; i <= m; i++) {
			dp[i][0][0] = 1;
		}
		double[][] c = new double[n + 2][n + 2];
		c[0][0] = 1;
		for (int i = 1; i < c.length; i++) {
			c[i][0] = 1;
			for (int j = 1; j <= n + 1; j++) {
				for (int k = 1; k <= i; k++) {
					c[i][j] += c[k][j - 1];
				}
			}
		}
		for (int i = 1; i <= m; i++) {
			for (int j = 1; j <= n; j++) {
				for (int k = 0; k <= n; k++) {
					// if (k > 0)
					// dp[i][j][k] += dp[i][j - 1][k];
					for (int t = 0; t <= j; t++) {
						int e = Math.max(k, (t + a[i - 1] - 1) / a[i - 1]);
						if (e > 0)
							dp[i][j][e] += dp[i - 1][j - t][k]
									* c[j - t + 1][t];
					}
				}
			}
		}
		double ans = 0;
		double ans2 = 0;
		for (int i = 0; i <= n; i++) {
			ans += (dp[m][n][i] /*- dp[m][n - 1][i]*/) * i;
			ans2 += dp[m][n][i];
		}
		out.printf("%.15f\n", ans / ans2);
	}
}