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
input = sys.stdin.readline

#input
N, M = map(int, input().split())
u = [[] for i in range(N)]
#（便宜的に地点から-1としておく）
for i in range(M):
    s, t = map(int, input().split())
    u[s-1].append(t-1)

#output
#地点vにたどり着く確率をp[v]とする。
p = [0] * N
p[0] = 1
for v in range(N):
    for i in range(v):
        if v in u[i]:
            p[v] += p[i]/len(u[i])

#期待値をゴールから逆順に求めていく。点vからゴールまでの移動距離の期待値をq[v]とする。
q = [0] * N
q[N-1] = 0
for v in reversed(range(N)):
    for i in u[v]:
        q[v] += (1 + q[i])/len(u[v])

v = [0] * N
for i in range(N-1):
    a = len(u[i])
    if a > 1:
        b = max(q[c] for c in u[i])
        v[i] = p[i]*(q[i] - (1+b))/(a-1)

print(q[0]+min(v))