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 "bits/stdc++.h"
using namespace std;
#define REP(i, n) for(ll i = 0;i < n;i++)
#define ll long long
#define MOD 1000000007LL
#define llMAX 9223372036854775807LL
#define llMIN -9223372036854775808LL
using vi = vector<ll>; // llの1次元の型に vi という別名をつける
using vvi = vector<vi>; // llの2次元の型に vvi という別名をつける
 
ll n;
ll M=1e5+1;
 
//bit全探索
vi bt;
void setbit(ll i){
  bt.clear();
  for(ll tmp=0;i>0;i/=2){
    if (i%2==1) bt.push_back(tmp);
    tmp++;
  }
}
 
 
 
 
int main(){
  ll ans=0,q;
  cin >>n;
  n++;
  ll nlg=ceil(log2(n));
  vi V(n),W(n);
  REP(i,n-1) cin>>V[i+1]>>W[i+1];
 
  cin >>q;
  vi u(q),L(q);
  REP(i,q) cin>>u[i]>>L[i];
 
  //上半分
  ll nhi=nlg-(nlg/2);
  if (nhi==9)nhi=10;
  vvi data(1LL<<nhi, vi(M));  // N * M の2次元配列
 
  for(ll i=1;i<(1LL<<nhi);i++){
    for(ll j=0;j<M;j++){
      if ((j-W[i])>=0)
      {
        data[i][j]=max(data[i/2][j-W[i]]+V[i],data[i/2][j]);
      }
      else
      {
        data[i][j]=data[i/2][j];
      }
    }
  }
 
 
ll uiv[10];
ll wuiv=0;
 
 
//下半分
for(ll loop=0;loop<q;loop++){
  ll uidx=u[loop];
  if (uidx<(1LL<<nhi)){
    cout<<data[uidx][L[loop]]<<endl;
  }else{//上＋下
    ll maxval=0;
    //vi uiv;
    ll nn=0;
    wuiv=0;
    for(;;uidx/=2){
      if (uidx<(1LL<<nhi))break;
      uiv[wuiv]=uidx;
      wuiv++;
      nn++;
    }
 
    ll bn=1<<nn;
    for(ll i=0;i<bn;i++){
      setbit(i);
      ll tmpw=0;
      ll tmpv=0;
      /*
      for(ll j=0;j<nn;j++){
        if (i&(1<<j)){
          tmpv+=V[uiv[j]];
          tmpw+=W[uiv[j]];
        }
      }
      */
      for (auto j : bt) {
        tmpv+=V[uiv[j]];
        tmpw+=W[uiv[j]];
      }
      if (tmpw<=L[loop]){
        maxval=max(data[uidx][L[loop]-tmpw]+tmpv,maxval);
      }
    }
 
    cout<<maxval<<endl;
  }
 
 
 
 
 
 
 
 
 
 
 
 
}
 
 
  return 0;
}