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:
ii = lambda : int(input())
mi = lambda : map(int,input().split())
li = lambda : list(map(int,input().split()))

def main():
    n,m = mi()
    tree = [[] for _ in range(n)]
    for i in range(m):
        s,t = mi()
        tree[s-1].append(t-1)
    dp = [0]*n
    dpl = [0]*n
    for i in range(n-1):
        dpl[i] = len(tree[i])
    for i in reversed(range(n-1)):
        for k in tree[i]:
            dp[i] += dp[k] 
        dp[i] /= dpl[i]
        dp[i] += 1
    ans = dp[0]
    
    def func(ss,tt):
        tmp = 0
        for l in tree[ss]:
            if l == tt:
                continue
            tmp += dp[l]
        tmp = tmp/(dpl[ss] -1) + 1
        if tmp > dp[ss]:
            return dp[0]
        else:
            dp[ss] = tmp
        
        for a in reversed(range(ss)):
            tmp = 0
            for b in tree[a]:
                tmp += dp[b]
            dp[a] = tmp/dpl[a] + 1
        
        return dp[0]
    
    for i in range(n-1):
        if len(tree[i]) == 1:
            continue
        for k in tree[i]:
            ans = min(ans,func(i,k))
    print(ans)

main()