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())
edge_out = [[] for i in range(N)] #edge_out[i]はiから出ている道
edge_in = [[] for i in range(N + 1)] # edge_in[i]はiに行く道
for i in range(M):
    s, t = map(int, input().split())
    edge_out[s].append(t)
    edge_in[t].append(s)

edge_sum = [len(edge_out[i]) for i in range(N)] # edge_sum[i] = iから出ている道の本数

dp = [0.0]*(N + 1) # dp[i]はiからNへの距離の期待値
dp[N - 1] = 1.0
# 道を塞がない場合
for i in range(N - 2, 0, - 1):
    dp[i] = 1 + sum(dp[j] for j in edge_out[i])/edge_sum[i]
    
p_to_i = [0]*(N + 1) # p_to_i[i]は1から各iへ到達する確率
p_to_i[1] = 1
for i in range(2, N + 1):
    p_to_i[i] = sum(p_to_i[j]/edge_sum[j] for j in edge_in[i])


t = 0

for i in range(1, N - 1):
    if edge_sum[i] == 1:
        continue
        # iから出ている道の先にある部屋のうち、その部屋からNまでの距離の期待値が最も大きいものへ通ずる道を塞ぐ
    t = max(t, p_to_i[i]*(dp[i] - 1- ((dp[i] - 1)*edge_sum[i] - max(dp[j] for j in edge_out[i]))/(edge_sum[i] - 1)))
    
print(dp[1] - t)