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>
#define st first
#define nd second

using ll = long long ;
using namespace std;
using pii = pair<int,int>;
const int N = (1 << 18) + 10;
int ans[(int)1e5 + 10];
vector<pii> ele[N], query[N];;
pii p[N];
bool sort_by_size(pii &lhs, pii &rhs){
	return lhs.nd < rhs.nd;
}
void solve(int id){
	if(id == 1){
		for(auto it: query[id]) 
			ans[it.nd] = (it.st <= p[1].nd) ? p[1].st : 0;
	}
	int mid = ((int)(ele[id].size())) >> 1, sz = (int)(ele[id].size());
	int ma = 0;
	for(auto it: query[id]){
		ma = max(ma, it.st);
	}
	vector<pii> v[2];
	for(int i = 1; i < (1 << mid); i++){
		pii val = {0, 0};
		for(int j = 0; j < mid; j++){
			if((1 << j) & i){
				val.st += ele[id][j].st, val.nd += ele[id][j].nd;
			}
		}
		if(val.nd <= ma) v[0].push_back(val);
	}
	v[0].push_back({0,0});
	for(int i = 1; i < (1 << (sz - mid)); i++){
		pii val = {0, 0};
		for(int j = 0; j < (sz - mid); j++){
			if((1 << j) & i){
				val.st += ele[id][j + mid].st, val.nd += ele[id][j + mid].nd;
			}
		}
		if(val.nd <= ma) v[1].push_back(val);
	}
	sort(v[0].begin(), v[0].end(), sort_by_size), sort(v[1].begin(), v[1].end(), sort_by_size);
	//printf("at %d\n", id);
	//for(auto it: v[0]) printf("{%d, %d} ", it.st, it.nd);
	//cout << '\n';
	//for(auto it: v[1]) printf("{%d, %d} ", it.st, it.nd);
	//cout << '\n';
	for(auto it: query[id]){
		ma = 0;
		int ma_2 = 0;
		int it1 = (int)(v[0].size()) - 1, it2 = 0;
		for(; it1 >= 0; it1 --){
			if(v[0][it1].nd >= it.st) continue;
			ma = max(ma, v[0][it1].st);
			for(; it2 < (int)v[1].size(); it2++){
				//printf("it2 = %d\n", it2);
				///printf("%d vs %d\n", v[0][it1].nd + v[1][it2].nd, it.st);
				if(v[0][it1].nd + v[1][it2].nd > it.st) break;
				ma_2 = max(ma_2, v[1][it2].st);
				
				//printf("ma_2 = %d\n", ma_2);
			}
			ma = max(ma, v[0][it1].st + ma_2);
		}
		ans[it.nd] = ma;
	}

}
void push_all(int id){
	if(!ele[id].empty()) return;
	push_all(id >> 1);
	for(auto it: ele[id >> 1]){
		ele[id].push_back(it);
	}
	ele[id].push_back(p[id]);
}
int main(){
	int n, q;
	scanf("%d", &n);
	for(int i =1; i <= n; i++){
		scanf("%d %d", &p[i].st, &p[i].nd);
	}
	ele[1].push_back(p[1]);
	scanf("%d", &q);
	for(int i = 1; i <= q; i ++){
		int x, y;
		scanf("%d %d", &x, &y);
		query[x].push_back({y, i});
	}
	for(int i = 1; i <= n; i++){
		if(query[i].empty()) continue;
		push_all(i);
		solve(i);
	}
	for(int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}