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())
a = [[] for i in range(n-1)]
for i in range(m):
  s, t = map(int, input().split())
  a[s-1].append(t-1)
dp = [0]*n
sum_d = [0]*(n-1)
max_d = [0]*(n-1)
b = [0]*n
for i in range(n-2,-1,-1):
  for j in a[i]:
    sum_d[i] += dp[j] + 1
    b[i] += 1
    if max_d[i] < dp[j]+1 and len(a[i]) > 1:
      max_d[i] = dp[j]+1  
  dp[i] = sum_d[i] / b[i]
min_dp = dp[0]
for k in range(n-2):
  if b[k]==1:
    continue
  ddp = [0]*k + [dp[k] - (sum_d[k]-max_d[k])/(b[k]-1)]
  for i in range(k-1,-1,-1):
    ddp[i] = sum([ddp[j] for j in a[i] if j <= k]) / b[i]
  min_dp = min(min_dp,dp[0] - ddp[0])
print(min_dp)