Problem:
Score : 700 points

Problem Statement

We have a rooted binary tree with N vertices, where the vertices are numbered 1 to N.
Vertex 1 is the root, and the parent of Vertex i (i \geq 2) is Vertex \left[ \frac{i}{2} \right].

Each vertex has one item in it. The item in Vertex i has a value of V_i and a weight of W_i.
Now, process the following query Q times:

Given are a vertex v of the tree and a positive integer L.
Let us choose some (possibly none) of the items in v and the ancestors of v so that their total weight is at most L.
Find the maximum possible total value of the chosen items.

Here, Vertex u is said to be an ancestor of Vertex v when u is an indirect parent of v, that is, there exists a sequence of vertices w_1,w_2,\ldots,w_k (k\geq 2) where w_1=v, w_k=u, and w_{i+1} is the parent of w_i for each i.

Constraints

All values in input are integers.

1 \leq N < 2^{18}

1 \leq Q \leq 10^5

1 \leq V_i \leq 10^5

1 \leq W_i \leq 10^5

For the values v and L given in each query, 1 \leq v \leq N and 1 \leq L \leq 10^5.

Input

Let v_i and L_i be the values v and L given in the i-th query.
Then, Input is given from Standard Input in the following format:

N
V_1 W_1
:
V_N W_N
Q
v_1 L_1
:
v_Q L_Q

Output

For each integer i from 1 through Q,
the i-th line should contain the response to the i-th query.

Sample Input 1

3
1 2
2 3
3 4
3
1 1
2 5
3 5

Sample Output 1

0
3
3

In the first query, we are given only one choice: the item with (V, W)=(1,2). Since L = 1, we cannot actually choose it, so our response should be 0.

In the second query, we are given two choices: the items with (V, W)=(1,2) and (V, W)=(2,3). Since L = 5, we can choose both of them, so our response should be 3.

Sample Input 2

15
123 119
129 120
132 112
126 109
118 103
115 109
102 100
130 120
105 105
132 115
104 102
107 107
127 116
121 104
121 115
8
8 234
9 244
10 226
11 227
12 240
13 237
14 206
15 227

Sample Output 2

256
255
250
247
255
259
223
253

Solution in Julia (1.4.0) (judge verdict: Runtime Error)

parseInt(x)=parse(Int, x)
parseMap(x::Array{SubString{String},1})=map(parseInt, x)

function bsearch(a::Array{Tuple{Int,Int},1},x)
	l,r = 0, length(a)+1
	while r-l>1
		m = (l+r)>>1
		if a[m][1]<=x
			l = m
		else
			r = m
		end
	end
	return(l)
end

function main()
	n = readline() |> parseInt
	a = Tuple{Int,Int}[(0,0) for i in 1:n]
	for i in 1:n
		v,w = readline() |> split |> parseMap
		a[i] = (v,w)
	end
	dps = [Dict{Int,Int}() for i in 1:n>>1]
	dps[1][0]=0
	dps[1][a[1][2]]=a[1][1]
	for iter in 2:n>>1
		v = iter
		dp = copy(dps[v>>1])
		(p,q)=a[v]
		for j in keys(dps[v>>1])
			if haskey(dp,j+q)
				dp[j+q]=max(dp[v>>1][j+q],dps[v>>1][j]+p)
			else
				dp[j+q]=dps[v>>1][j]+p
			end
		end
		dps[v] = copy(dp)
	end
	dpt = [Tuple{Int,Int}[] for i in 1:n>>1]
	for i in 1:n>>1
		y = Tuple{Int,Int}[(j,dps[i][j]) for j in keys(dps[i])]
		y = sort(y,by=x->x[1])
		z = Tuple{Int,Int}[]
		push!(z,y[1])
		for j in 2:length(y)
			push!(z,(y[j][1],max(y[j][2],z[j-1][2])))
		end
		dpt[i]=copy(z)
	end
	q = readline() |> parseInt
	for iter in 1:q
		v,l = readline() |> split |> parseMap
		dp = Dict{Int,Int}()
		dp[0]=0
		while v>n>>1
			(p,q) = a[v]
			tmp = copy(dp)
			tmp[0]=0
			for j in keys(dp)
				if j+q<=l
					if haskey(dp,j+q)
						tmp[j+q]=max(dp[j+q],dp[j]+p)
					else
						tmp[j+q]=dp[j]+p
					end
				end
			end
			dp = copy(tmp)
			v>>=1
		end
		mx = 0
		for j in keys(dp)
			k = bsearch(dpt[v],l-j)
			if k>0
				mx = max(mx,dp[j]+dpt[v][k][2])
			else
				mx = max(mx,dp[j])
			end
		end
		println(mx)
	end
end

main()