Allen and Bessie are playing a simple number game. They both know a function $f: \{0, 1\}^n \to \mathbb{R}$, i. e. the function takes $n$ binary arguments and returns a real value. At the start of the game, the variables $x_1, x_2, \dots, x_n$ are all set to $-1$. Each round, with equal probability, one of Allen or Bessie gets to make a move. A move consists of picking an $i$ such that $x_i = -1$ and either setting $x_i \to 0$ or $x_i \to 1$.
After $n$ rounds all variables are set, and the game value resolves to $f(x_1, x_2, \dots, x_n)$. Allen wants to maximize the game value, and Bessie wants to minimize it.
Your goal is to help Allen and Bessie find the expected game value! They will play $r+1$ times though, so between each game, exactly one value of $f$ changes. In other words, between rounds $i$ and $i+1$ for $1 \le i \le r$, $f(z_1, \dots, z_n) \to g_i$ for some $(z_1, \dots, z_n) \in \{0, 1\}^n$. You are to find the expected game value in the beginning and after each change.
-----Input-----
The first line contains two integers $n$ and $r$ ($1 \le n \le 18$, $0 \le r \le 2^{18}$).
The next line contains $2^n$ integers $c_0, c_1, \dots, c_{2^n-1}$ ($0 \le c_i \le 10^9$), denoting the initial values of $f$. More specifically, $f(x_0, x_1, \dots, x_{n-1}) = c_x$, if $x = \overline{x_{n-1} \ldots x_0}$ in binary.
Each of the next $r$ lines contains two integers $z$ and $g$ ($0 \le z \le 2^n - 1$, $0 \le g \le 10^9$). If $z = \overline{z_{n-1} \dots z_0}$ in binary, then this means to set $f(z_0, \dots, z_{n-1}) \to g$.
-----Output-----
Print $r+1$ lines, the $i$-th of which denotes the value of the game $f$ during the $i$-th round. Your answer must have absolute or relative error within $10^{-6}$.
Formally, let your answer be $a$, and the jury's answer be $b$. Your answer is considered correct if $\frac{|a - b|}{\max{(1, |b|)}} \le 10^{-6}$.
-----Examples-----
Input
2 2
0 1 2 3
2 5
0 4
Output
1.500000
2.250000
3.250000
Input
1 0
2 3
Output
2.500000
Input
2 0
1 1 1 1
Output
1.000000
-----Note-----
Consider the second test case. If Allen goes first, he will set $x_1 \to 1$, so the final value will be $3$. If Bessie goes first, then she will set $x_1 \to 0$ so the final value will be $2$. Thus the answer is $2.5$.
In the third test case, the game value will always be $1$ regardless of Allen and Bessie's play.
R = lambda:list(map(int,input().split()))
sum1 = 0
p1 = 1
a = [0]*263000
b = a
n,r = R()
p1 = 2**n
a = R()
sum1 = sum(a)
p2 = sum1 / p1
for i in range(r):
	z,g=R()
	sum1 = sum1 - a[z] + g
	a[z] = g
	b[i] = sum1/p1
print(p2)
for i in range(r):
	print(b[i])
R = lambda:list(map(int,input().split()))
sum1 = 0
p1 = 1
a = [0]*263000
b = a
n,r = R()
p1 = 2**n
a = R()
sum1 = sum(a)
p2 = sum1 / p1
for i in range(r):
	z,g=R()
	sum1 = sum1 - a[z] + g
	a[z] = g
	b[i] = sum1/p1
print(p2)
for i in range(r):
	print(b[i])