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:

N,M = map(int,input().split())

lis = [[] for i in range(N)]
rlis = [[] for i in range(N)]
mlis = [0] * N

for i in range(M):

    s,t = map(int,input().split())
    s -= 1
    t -= 1

    lis[s].append(t)
    rlis[t].append(s)


odp = [0] * N

for i in range(N):

    i = N - 1 - i

    for j in lis[i]:

        odp[i] += (odp[j] + 1)/len(lis[i])
        

ans = odp[0]


for k in range(N):

    dp = [0] * N
    flag = True

    for i in range(N):

        i = N - 1 - i

        if i != k:

            if k < i:
                dp[i] = odp[i]

            else:

                for j in lis[i]:

                    dp[i] += (dp[j] + 1)/len(lis[i])

        else:

            maxi = 0

            for j in lis[i]:

                maxi = max(maxi,dp[j])
                dp[i] += (dp[j] + 1)

            if len(lis[i]) > 1:
                dp[i] = (dp[i]-(maxi+1)) / (len(lis[i])-1)

            else:
                flag = False
                
    #print (k,dp)

    if flag:
        ans = min(ans,dp[0])

print (ans)