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;

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	long long n;
	cin >> n;
	vector<int> vs(n);
	vector<long long> ws(n);
	for(int i=0;i<n;i++){
		cin >> vs[i] >> ws[i];
	}
	int q;
	cin >> q;
	
	vector<vector<pair<int,int>>> rek(n);
	int han=9;
	for(int i=0;i<n;i++){
		vector<int> ta,ta2;
		int cur=i;
		for(int j=0;j<han;j++){
			ta.push_back(ws[cur]);
			ta2.push_back(vs[cur]);
			int nex=(cur-1)/2;
			if(cur==0) break;
			cur=nex;
		}
		
		vector<pair<int,int>> zen(1,make_pair(0,0));
		for(int j=0;j<(int)ta.size();j++){
			vector<pair<int,int>> nzen=zen;
			for(int k=0;k<(int)zen.size();k++){
				pair<int,int> p0=make_pair(zen[k].first+ta[j],zen[k].second+ta2[j]);
				nzen.push_back(p0);
			}
			nzen.swap(zen);
		}
		sort(zen.begin(),zen.end());
		int mx=0;
		for(int i=0;i<(int)zen.size();i++){
			mx=max(mx,zen[i].second);
			zen[i].second=mx;
			//cout << zen[i].first << " " << zen[i].second << "\n";
		}
		//cout << "\n";
		rek[i]=zen;
	}
	
	for(int tc=0;tc<q;tc++){
		int v,L;
		cin >> v >> L;
		v--;
		
		vector<int> left;
		for(int i=0;i<(int)rek[v].size();i++){
			left.push_back(rek[v][i].first);
		}
		int cur=v;
		bool bo=true;
		for(int j=0;j<han;j++){
			int nex=(cur-1)/2;
			if(cur==0){
				bo=false;
				break;
			}
			cur=nex;
		}
		if(!bo){
			int ind=(int)(upper_bound(left.begin(),left.end(),L)-left.begin());
			if(ind==0){
				cout << 0 << "\n";
			}else{
				ind--;
				cout << rek[v][ind].second << "\n";
			}
			continue;
		}
		vector<int> right;
		for(int i=0;i<(int)rek[cur].size();i++){
			right.push_back(rek[cur][i].first);
		}
		
		int res=0;
		for(int i=0;i<(int)left.size();i++){
			if(left[i]>L) break;
			int ind=(int)(upper_bound(right.begin(),right.end(),L-left[i])-right.begin());
			if(ind==0){
				
			}else{
				ind--;
				res=max(res,rek[v][i].second+rek[cur][ind].second);
			}
		}
		cout << res << "\n";
	}
	return 0;
}