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 int64_t llo;
#define mp make_pair
#define pb push_back
#define a first
#define b second
#define endl '\n'
int n;
int aa[300000];
int bb[300000];
int q;
int no,w;
vector<int> cc;
vector<int> dd;
vector<pair<int,int>> kk;
vector<pair<int,int>> mm;
void brute(int ind ,int su=0,int val=0){
	if(ind==cc.size()){
		if(su<=w){
			kk.pb({su,val});
		}
	}
	else{
		brute(ind+1,su+bb[cc[ind]-1],val+aa[cc[ind]-1]);
		brute(ind+1,su,val);
	}
}
void brute2(int ind ,int su=0,int val=0){
	if(ind==dd.size()){
		if(su<=w){
			mm.pb({su,val});
		}		
	}
	else{
		brute2(ind+1,su+bb[dd[ind]-1],val+aa[dd[ind]-1]);
		brute2(ind+1,su,val);
	}
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>aa[i]>>bb[i];
	}
	cin>>q;
	while(q--){
		cc.clear();
		dd.clear();
		kk.clear();
		mm.clear();
		cin>>no>>w;
		if(no==1){
			if(bb[0]<=w){
				cout<<aa[0]<<endl;
			}
			else{
				cout<<0<<endl;
			}
			continue;
		}
		
		while(no>0){
			cc.pb(no);
			no/=2;
		
		}
		int x=(cc.size())/2;
		while(cc.size()>x){
			dd.pb(cc.back());
			cc.pop_back();
		}
		brute(0);
		brute2(0);
		sort(kk.begin(),kk.end());
		sort(mm.begin(),mm.end());
		int ma=0;
	/*	for(auto j:kk){
			cout<<j.a<<","<<j.b<<endl;
		}
		for(auto j:mm){
			cout<<j.a<<":"<<j.b<<endl;
		}
		cout<<endl<<endl;*/
		for(int i=0;i<kk.size();i++){
			kk[i].b=max(kk[i].b,ma);
			ma=max(ma,kk[i].b);
		}
		ma=0;
		for(int i=0;i<mm.size();i++){
			mm[i].b=max(mm[i].b,ma);
			ma=max(ma,mm[i].b);
		}
		/*for(auto j:kk){
			cout<<j.a<<","<<j.b<<endl;
		}
		for(auto j:mm){
			cout<<j.a<<":"<<j.b<<endl;
		}*/
		int ans=0;
		int ind=mm.size()-1;
		for(int i=0;i<kk.size();i++){
			while(ind>=0){
				if(mm[ind].a+kk[i].a>w){
					ind--;
				}
				else{
					break;
				}
			}
			if(ind>=0){
				ans=max(ans,kk[i].b+mm[ind].b);
			}
		}
		cout<<ans<<endl;
	}






	return 0;
}