Problem:
Score : 600 points

Problem Statement

There is a cave consisting of N rooms and M one-directional passages. The rooms are numbered 1 through N.

Takahashi is now in Room 1, and Room N has the exit. The i-th passage connects Room s_i and Room t_i (s_i < t_i) and can only be traversed in the direction from Room s_i to Room t_i. It is known that, for each room except Room N, there is at least one passage going from that room.

Takahashi will escape from the cave. Each time he reaches a room (assume that he has reached Room 1 at the beginning), he will choose a passage uniformly at random from the ones going from that room and take that passage.

Aoki, a friend of Takahashi's, can block one of the passages (or do nothing) before Takahashi leaves Room 1. However, it is not allowed to block a passage so that Takahashi is potentially unable to reach Room N.

Let E be the expected number of passages Takahashi takes before he reaches Room N. Find the value of E when Aoki makes a choice that minimizes E.

Constraints

2 \leq N \leq 600

N-1 \leq M \leq \frac{N(N-1)}{2}

s_i < t_i

If i != j, (s_i, t_i) \neq (s_j, t_j). (Added 21:23 JST)

For every v = 1, 2, ..., N-1, there exists i such that v = s_i.

Input

Input is given from Standard Input in the following format:

N M
s_1 t_1
:
s_M t_M

Output

Print the value of E when Aoki makes a choice that minimizes E.
Your output will be judged as correct when the absolute or relative error from the judge's output is at most 10^{-6}.

Sample Input 1

4 6
1 4
2 3
1 3
1 2
3 4
2 4

Sample Output 1

1.5000000000

If Aoki blocks the passage from Room 1 to Room 2, Takahashi will go along the path 1 → 3 → 4 with probability \frac{1}{2} and 1 → 4 with probability \frac{1}{2}. E = 1.5 here, and this is the minimum possible value of E.

Sample Input 2

3 2
1 2
2 3

Sample Output 2

2.0000000000

Blocking any one passage makes Takahashi unable to reach Room N, so Aoki cannot block a passage.

Sample Input 3

10 33
3 7
5 10
8 9
1 10
4 6
2 5
1 7
6 10
1 4
1 3
8 10
1 5
2 6
6 9
5 6
5 8
3 6
4 8
2 7
2 9
6 7
1 2
5 9
6 8
9 10
3 9
7 8
4 5
2 10
5 7
3 5
4 7
4 9

Sample Output 3

3.0133333333

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.PrintStream;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
	static final PrintStream out = System.out;
	static final Scanner sc = new Scanner(System.in);
	static final long Mod = 1000_000_007L;
	List<Integer>[] paths;
	double[] dp = new double[650];
	int N,M;
	double res = 1<<60;

	public static void main(String[] args) {
		Main main = new Main();
		main.solve();
		System.exit(0);
	}

	void solve() {
		N = nextInt();
		M = nextInt();
		paths = new List[N+1];
        for (int i=0;i<=N;i++) {
            paths[i] = new ArrayList<>();
        }

		for(int i=1;i<=M;i++) {
			int s,t;
			s = nextInt();
			t = nextInt();
			paths[s].add(t);
		}
	    for(int i=0;i<N;i++) {
	        f(i);
	    }
	    out.println(res);
	}

	void f(int block) {
	    dp[N] = 0;
	    for(int i=N-1;i>=1;i--) {
	        double sum = 0.0;
	        double maxi = 0.0;
	        int num = paths[i].size();
	        for(Integer to : paths[i]) {
	        	out.println(to);
	            sum += dp[to];
	            maxi = Math.max(maxi, dp[to]);
	        }
	        if(i == block && num >= 2) {
	            num--;
	            sum -= maxi;
	        }
	        dp[i] = sum / num + 1;
	    }
	    res = Math.min(res, dp[1]);
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)a[i] = nextInt();
		return a;
	}
	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)a[i] = nextLong();
		return a;
	}
	public int nextInt() {
		return Integer.parseInt(next());
	}
	public long nextLong() {
		return Long.parseLong(next());
	}
	public String next() {
		return sc.next();
	}

	long gcd(long m,long n) {
		if (m<n) return gcd(n,m);
		if (n==0)return m;
		return gcd(n,m%n);
	}
}