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:
#s<t!
#一番期待値が悪くなる辺だけ考える
n,m = map(int, input().split( ))

E= [[] for _ in range(n)]

for _ in range(m):
    s,t = map(int, input().split( ))
    E[s-1].append(t-1)

dp = [0]*n 

for s in range(n-2,-1,-1):
    for t in E[s]:
        dp[s] += dp[t]
    if E[s]:
        dp[s] /= len(E[s])
    dp[s] += 1

#print("dp",*dp)

ans = dp[0]
for s2 in range(n-1):
    if len(E[s2])-1:
        dp2 = [0]*n
        mxi = 0
        mxt = -1
        for t in E[s2]:
            if mxt<dp[t]:
                mxi = t 
                mxt = dp[t]
        if mxt<0:
            continue            
        for s in range(n-2,-1,-1):
            for t in E[s]:
                dp2[s] += dp2[t]
            if s ==s2:###ループ深さ
                dp2[s] -= mxt
            if s == s2:
                dp2[s] /= (len(E[s])-1)
            elif E[s]:##el if
                dp2[s] /= len(E[s])
            dp2[s] += 1
        ans = min(ans, dp2[0])
        #print("dp2",*dp2)

print(ans)