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: Time Limit Exceeded)

#include <iostream>
#include <math.h>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;
long long knap[1024][100001];
int main(){
  int N;
  cin>>N;
  long long V[1000000],W[1000000];
  for(int i=1;i<=N;i++){
    cin>>V[i]>>W[i];
  }
  for(int i=0;i<=min(511,N);i++){
    for(int j=0;j<=100000;j++){
      if(j==0) knap[i][j]=0;
      else if(i==0){
        knap[i][j]=0;
      }
      else{
        if(j-W[i]>=0) knap[i][j]=max(knap[i/2][j],knap[i/2][j-W[i]]+V[i]);
        else knap[i][j]=knap[i/2][j];
      }
    }
  }
  int Q;
  cin>>Q;
  for(int i=0;i<Q;i++){
    int v,L;
    cin>>v>>L;
    if(v>=512){
    long long k=1;
    int u=v;
    while(u>=512){
      k=k*2;
      u=u/2;
    }
    long long answer=0;
    for(int a=0;a<k;a++){
      long long val=0;
      long long wei=0;
      long long K=1;
      int J=v;
      while(J>u){
        if((a/K)%2==1){
          val+=V[J];
          wei+=W[J];
        }
        J=J/2;
        K=K*2;
      }
      if(wei<=L&&answer<val+knap[u][L-wei]) answer=val+knap[u][L-wei];
    }
    cout<<answer<<endl;
    }
    else cout<<knap[v][L]<<endl;
  }
  return 0;
}