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 math

N, M = [int(x) for x in input().split()]
edges = [[int(x) for x in input().split()] for _ in range(M)]

forward_edges = {n+1:[] for n in range(N)}
reverse_edges = {n+1:[] for n in range(N)}
for e in edges:
  forward_edges[e[0]].append(e[1])
  reverse_edges[e[1]].append(e[0])
# print(  forward_edges)
# print(  reverse_edges)
estimated_distance = {n+1:None for n in range(N)}
estimated_distance[N] = 0
probability = {n+1:None for n in range(N)}
probability[1] = 1
def get_probability(n):
  if probability[n] is not None:
    return probability[n]
  prob = 0
  for m in reverse_edges[n]:
    prob += get_probability(m) / len(forward_edges[m])
  probability[n] = prob
  return prob
  
def get_distance(n):
  if estimated_distance[n] is not None:
    return estimated_distance[n]
  total_distance = 0
  for m in forward_edges[n]:
    total_distance += get_distance(m)
  estimated_distance[n] = 1 + total_distance / len(forward_edges[n])
  return estimated_distance[n]


before_distance = get_distance(1)
max_decrease = 0
for n in range(1, N):
  if len(forward_edges[n]) == 1:
    continue
  distances = [get_distance(m) for m in forward_edges[n]]
  distance_diff = sum(distances) / len(distances) - (
    sum(distances)-max(distances))/(len(distances)-1)
  decrease = get_probability(n) * distance_diff
  max_decrease = max(max_decrease, decrease)

print(before_distance - max_decrease)