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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

# reference: https://www.onakasuitacity.com/abc144-f/

import numpy as np

n, m = map(int, input().split())
outgoing = [[] for _ in range(n)]
incoming = [[] for _ in range(n)]
for _ in range(m):
    si, ti = map(int, input().split())
    outgoing[si-1].append(ti-1)
    incoming[ti-1].append(si-1)


ex_dp = np.zeros(n)
for v in reversed(range(n-1)):   # O(m)
    outgoing_degree = len(outgoing[v])
    ex_dp[v] = 1 + sum(ex_dp[post] for post in outgoing[v])/outgoing_degree if outgoing_degree!=0 else np.inf

p_dp = np.ones(n)
for v in range(1,n):      # O(m)
    p_dp[v] = sum(p_dp[prev]/len(outgoing[prev]) for prev in incoming[v])

min_ex = ex_dp[0]
for v, posts in enumerate(outgoing):      # O(m)
    for w in posts:     # cut: edge from v to w
        outgoing_degree = len(outgoing[v])
        if outgoing_degree <= 1:
            continue
        diff_remain_edge_p = 1/(outgoing_degree*(outgoing_degree-1))
        diff_cut_edge_p = 1/outgoing_degree
        diff_v_ex = sum(diff_remain_edge_p * ex_dp[post] for post in outgoing[v] if post!=w) - diff_cut_edge_p * ex_dp[w]
        min_ex = min(ex_dp[0] + p_dp[v] * diff_v_ex, min_ex)

print(min_ex)