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

n, m = map(int, input().split())
links = [set() for _ in range(n)]
counts = [0] * n
for line in sys.stdin:
    s, t = map(int, line.split())
    s -= 1
    t -= 1
    links[s].add(t)
    counts[s] += 1

# Expected number of edges passing from i to N
expected = [0.] * n

# i から伸びる辺を1本消すことで削減できる"iからNまでの辺数の期待値"の最大量
reducible = [0.] * n

for i in range(n - 2, -1, -1):
    nxt_exp = [expected[j] for j in links[i]]
    sum_exp = sum(nxt_exp)
    cnt = counts[i]

    expected[i] = sum_exp / cnt + 1

    if cnt > 1:
        reduced_exp = (sum_exp - max(nxt_exp)) / (cnt - 1) + 1
        reducible[i] = expected[i] - reduced_exp
    else:
        reducible[i] = 0.

# Probability of visiting i when starting from 1
probability = [0.] * n
probability[0] = 1.

for i in range(n - 1):
    jp = probability[i] / counts[i]
    for j in links[i]:
        probability[j] += jp

print(expected[0] - max(p * r for p, r in zip(probability, reducible)))