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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int fa[1<<18],f[1<<9][N],v[1<<18],w[1<<18];
int main(){
	int n;cin>>n;
	for(int i=1;i<=n;++i)
		cin>>v[i]>>w[i],fa[i]=i/2;
	for(int i=1;i<(1<<9);++i){
		for(int j=1;j<=1e5;++j)
			f[i][j]=f[fa[i]][j];
		for(int j=w[i];j<=1e5;++j)
			f[i][j]=max(f[i][j],f[fa[i]][j-w[i]]+v[i]);
	}
	int q,x,l;
	cin>>q;
	while(q--){
		cin>>x>>l;
		if(x<(1<<9)){
			cout<<f[x][l]<<endl;
			continue;
		}
		vector< pair<int,int> > vec;
		while(x>=(1<<9)) vec.push_back(make_pair(v[x],w[x])),x=fa[x];
		int cnt=vec.size(),ans=f[x][l];
		for(int i=0;i<(1<<cnt);++i){
			int val=0,wei=0;
			for(int j=0;j<cnt;++j)
				if((i>>j)&1) val+=vec[j].first,wei+=vec[i].second;
			if(wei<=l) ans=max(ans,f[x][l-wei]+val);
		}
		cout<<ans<<endl;
	}
	return 0;
}