Problem: 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}.

Examples

Input

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


Output

1.5000000000


Input

3 2
1 2
2 3


Output

2.0000000000


Input

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


Output

3.0133333333

Solution (in PYTHON3):

Solution:
import sys
input = sys.stdin.readline

def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))

def main():
    import queue
    mod=10**9+7
    N,M=MI()
    adj=[[]for _ in range(N)]
    for i in range(M):
        s,t=MI()
        s-=1
        t-=1
        adj[s].append(t)
    
    """    
    逆向きに見ればO(M)，全ての通路で試すとO(M^2)
    切るルートをある程度まとめたい，i番目から出る通路のなかで一番でかいやつを無視する，というのをN回
    """
    
    ans=10**10
    
    for i in range(N-1):#i番目から出る辺を切る
        dp=[0]*N
        for j in range(N-2,-1,-1):#j番目の部屋から出るための期待値を求める．
            cnt=0#通り数
            if j!=i:
                for k in adj[j]:#jからkへ伸びる辺を考える
                    dp[j]+=dp[k]+1
                    cnt+=1
                dp[j]/=cnt
                    
            else:
                max_d=0
                for k in adj[j]:#jからkへ伸びる辺を考える
                    dp[j]+=dp[k]+1
                    cnt+=1
                    max_d=max(max_d,dp[k])
                
                #出る道が1本なら何もしない
                if cnt==1:
                    pass
                else:#一番でかいところをつぶす
                    cnt-=1
                    dp[j]-=(max_d+1)
                    dp[j]/=cnt
        # print(dp)
        if dp[0]!=0:
            ans=min(ans,dp[0])
            
    print(ans)
        

                
        
        
        
        
        
        
            
            
    
    

main()