p02884 AtCoder Beginner Contest 144 - Fork in the Road
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
```python
N, M = map(int, input().split())
G = [[] for _ in range(N)]
rG = [[] for _ in range(N)]
for _ in range(M):
    s, t = map(int, input().split())
    G[s-1].append(t-1)
E = [0]*N
for v in range(N-1)[::-1]:
    n = len(G[v])
    for nv in G[v]:
        E[v]+=(E[nv]+1)*(1/n)
ans = E[0]
for v in range(N-1):
    if len(G[v]) == 1:
        continue
    index = -1
    tmp = -1
    for i, nv in enumerate(G[v]):
        if tmp<E[nv]:
            index = nv
            tmp = E[nv]
    n = len(G[v])-1
    E[v] = 0
    for nv in G[v]:
        if nv == index:
            continue
        E[v]+=(E[nv]+1)*(1/n)
    for vv in range(v)[::-1]:
        E[vv] = 0
        n = len(G[vv])
        for nv in G[vv]:
            E[vv]+=(E[nv]+1)*(1/n)
    ans = min(ans, E[0])
print(ans)
```