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: Accepted)

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#include <utility>
#include <functional>
#include <time.h>
#include <stack>
#include <array>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;

int main()
{
	int n;
	cin>>n;
	int V[1<<18], W[1<<18];
	for(int i=1; i<=n; i++) cin>>V[i]>>W[i];
	const int d=10;
	vector<P> wv[1<<10];
	if(n>=(1<<d)){
		for(int t=(1<<(d-1)); t<(1<<d); t++){
			int v=t;
			vector<int> vs;
			while(v>1){
				vs.push_back(v);
				v>>=1;
			}
			vs.push_back(1);
			int m=vs.size();
			for(int i=0; i<(1<<m); i++){
				int vsum=0, wsum=0;
				for(int j=0; j<m; j++){
					if(i&(1<<j)){
						vsum+=V[vs[j]], wsum+=W[vs[j]];
					}
				}
				wv[t].push_back({wsum, vsum});
			}
			sort(wv[t].begin(), wv[t].end());
			for(int i=1; i<wv[t].size(); i++){
				wv[t][i].second=max(wv[t][i].second, wv[t][i-1].second);
			}
		}
	}
	int q; cin>>q;
	while(q--){
		int v, l;
		cin>>v>>l;
		if(v<(1<<d)){
			vector<int> vs;
			while(v>1){
				vs.push_back(v);
				v>>=1;
			}
			vs.push_back(1);
			int m=vs.size();
			int ans=0;
			for(int i=0; i<(1<<m); i++){
				int vsum=0, wsum=0;
				for(int j=0; j<m; j++){
					if(i&(1<<j)){
						vsum+=V[vs[j]], wsum+=W[vs[j]];
					}
				}
				if(wsum<=l) ans=max(ans, vsum);
			}
			cout<<ans<<endl;
			continue;
		}
		vector<int> vs;
		while(v>=(1<<d)){
			vs.push_back(v);
			v>>=1;
		}
		int m=vs.size();
		int ans=0;
		const int INF=1e9+7;
		for(int i=0; i<(1<<m); i++){
				int vsum=0, wsum=0;
				for(int j=0; j<m; j++){
					if(i&(1<<j)){
						vsum+=V[vs[j]], wsum+=W[vs[j]];
					}
				}
				if(l-wsum<0) continue;
			int t=lower_bound(wv[v].begin(), wv[v].end(), P(l-wsum, INF))-wv[v].begin();
			ans=max(ans, wv[v][t-1].second+vsum);
		}
		cout<<ans<<endl;
	}
	return 0;
}