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 Ruby (2.3.3) (judge verdict: Accepted)

N, M = gets.chomp.split(" ").map(&:to_i);
CAV = []
(N + 1).times{|i| CAV << []}
M.times{|i|
  f, t = gets.chomp.split(" ").map(&:to_i);
  CAV[f] << t
}

expect = Array.new(N + 1, 0.0)
BRN = Hash.new
(N + 1).times{|i| BRN[i] = CAV[i].size}

(N - 1).downto(1){|rom|
  expect[rom] = CAV[rom].inject(0){|rst, cav| rst += expect[cav]}.to_f / BRN[rom] + 1
}
dm = 0  #difference max
BRN.sort_by{|k, v| v}.each{|rom, branch|
  next if branch <= 1  #can't remove
  dif = Array.new(N + 1, 0.0)
  dif[rom] = (CAV[rom].map{|to| expect[to]}.max - expect[rom] + 1) / (branch - 1)
  next if dif[rom] < dm
  (rom - 1).downto(1){|i|
    dif[i] = CAV[i].inject(0){|rst, cav| rst += dif[cav]}.to_f / BRN[i]
  }
  dm = dif[1] if dif[1] > dm
}

puts expect[1] - dm