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
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

N,M = map(int,readline().split())
m = map(int,read().split())
ST = zip(m,m)

out_edge = [[] for _ in range(N+1)]
in_edge = [[] for _ in range(N+1)]

for s,t in ST:
    out_edge[s].append(t)
    in_edge[t].append(s)

out_deg = [len(x) for x in out_edge]

achieve_prob = [0]*(N+1)
achieve_prob[1] = 1.0
for n in range(2,N+1):
    achieve_prob[n] = sum(achieve_prob[p]/out_deg[p] for p in in_edge[n])

# その頂点に辿り着くまでの回数期待値
# いらなかったっぽい
# expected_to = [0]*(N+1)
# for n in range(2,N+1):
#     expected_to[n] = 1 + sum(expected_to[p]*achieve_prob[p]/out_deg[p] for p in in_edge[n])/achieve_prob[n]

# その頂点に辿り着いてからの回数期待値
expected_from = [0]*(N+1)
for n in range(N-1,0,-1):
    expected_from[n] = 1+sum(expected_from[c] for c in out_edge[n])/out_deg[n]

current_time = expected_from[1]

improve = 0
for n in range(1,N):
    if out_deg[n] == 1:
        continue
    x = expected_from[n]
    d = out_deg[n]
    y = (d*(x-1)-max(expected_from[c] for c in out_edge[n]))/(d-1) + 1
    value = (x-y)*achieve_prob[n]
    if improve < value:
        improve = value

answer = current_time - improve
print(answer)