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 PyPy2 (5.6.0) (judge verdict: Accepted)

# -*- coding: utf-8 -*-
import sys
from collections import deque

N,M=map(int, sys.stdin.readline().split())
al=[ [] for i in range(N+1) ]	#隣接リスト
rev_al=[ [] for i in range(N+1) ]	#リバース隣接リスト
O=[ 0 for i in range(N+1) ] #出次数
edge=[] #辺の集合

for _ in range(M):
    s,t=map(int, sys.stdin.readline().split())
    al[s].append(t)
    rev_al[t].append(s)
    #I[t]+=1
    O[s]+=1
    edge.append((s,t))


#頂点毎の到達する確立
P=[ 0 for i in range(N+1) ]   #確率
P[1]=1
 
for x in range(1,N+1):
    if O[x]!=0:
        p=float(1)/O[x]
        for to in al[x]:
            P[to]+=P[x]*p
        
#頂点毎の期待値
E=[ 0 for i in range(N+1) ]   #期待値

for x in range(N,0,-1):
    for to in rev_al[x]:
        p=float(1)/O[to]
        E[to]+=(E[x]+1)*p


#消す辺
ans=E[1]
for del_s,del_t in edge:
    if O[del_s]-1==0:
        continue
    new_E=float(E[del_s]*O[del_s]-(E[del_t]+1))/(O[del_s]-1)
    if new_E<E[del_s]:
        ans=min(ans,E[1]-(E[del_s]-new_E)*P[del_s])

print ans