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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
int n,Q;
#define Maxn 100010
int V[Maxn],W[Maxn];
int f[512][100010];
int seq1[Maxn],seq2[Maxn],cnt;
int to[Maxn];
int val1[Maxn],val2[Maxn];
void dfs(int u){
	if(u>=512)return;
	if(u>n)return;
		int fa=u/2;
		for(int i=0;i<=100000;++i){
			f[u][i]=f[fa][i];
			if(i>=W[u])f[u][i]=max(f[u][i],f[fa][i-W[u]]+V[u]);
		}
	dfs(u<<1);
	dfs(u<<1|1);
}

int lowbit(int x){return x&(-x);}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%d%d",&V[i],&W[i]);
	}
	dfs(1);
	scanf("%d",&Q);
	int u,L;
	for(int i=0;i<9;++i)to[(1<<i)]=i;
	while(Q--){
		scanf("%d%d",&u,&L);
		if(u<512)printf("%d\n",f[u][L]);
		else{
			cnt=0;
			while(u>=512){
				seq1[cnt]=V[u];
				seq2[cnt]=W[u];
				u>>=1;
				cnt++;
			}
			int Ans=f[u][L];
			for(int i=1;i<(1<<cnt);++i){
				int d=lowbit(i);
				val1[i]=val1[i^d]+seq1[to[d]];
				val2[i]=val2[i^d]+seq2[to[d]];
				if(val1[i]<=L)Ans=max(Ans,f[u][L-val1[i]]+val2[i]);
			}
			printf("%d\n",Ans);
		}
	}	
	return 0;
}