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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=300300;
int w[N],v[N];
int lg[N];
typedef pair<int,int> P;
typedef vector<P> VP;
VP qaq(VP a,int l){
	VP ps;
	ps.push_back(make_pair(0,0));
	for(P p:a){
		VP tmp;
		for(P tp:ps){
			if(tp.first+p.first<=l){
				tmp.push_back(P{tp.first+p.first,tp.second+p.second});
			}
			else{
				break;
			}
		}
		int w=ps.size();
		ps.insert(ps.end(),tmp.begin(),tmp.end());
		inplace_merge(ps.begin(),ps.begin()+w,ps.end());
		for(int i=0;i+1<(int)ps.size();i++){
			if(ps[i+1].second<=ps[i].second)ps[i+1]=ps[i];
		}
		for(int i=(int)ps.size()-1;i>=1;i--){
			if(ps[i].first==ps[i-1].first)ps[i-1]=ps[i];
		}
		ps.erase(unique(ps.begin(),ps.end()),ps.end());
	}
	return VP(ps.begin(),ps.end());
}
int solve(VP a,int l){
	int n=a.size();
	int mid=n/2;
	VP m1=qaq(VP(a.begin(),a.begin()+mid),l);
	VP m2=qaq(VP(a.begin()+mid,a.end()),l);
	int ans=0;
	VP::iterator it=--m2.end();
	for(P p:m1){
		while(it->first+p.first>l){
			it--;
		}
		ans=max(ans,p.second+it->second);
	}
	return ans;
}
int main(){
	for(int i=0;(1<<i)<N;i++)lg[1<<i]=i;
	ios::sync_with_stdio(false);
	int n;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>v[i]>>w[i];
	}
	int q;
	cin>>q;
	while(q--){
		int u,l;
		cin>>u>>l;
		VP a;
		while(u){
			a.push_back(P(w[u],v[u]));
			u>>=1;
		}
		cout<<solve(a,l)<<endl;
	}
	return 0;
}