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>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
typedef long double ld;
const int inf=1e9+7;
const ll longinf=1LL<<60;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define F first
#define S second
constexpr char ln = '\n';

const int mx=100010;
const ll mod=1e9+7;

#define bit(n,k) ((n>>k)&1) //*n no k bit me 1 or 0*/


int main(){
  int n;
  cin >> n;
  vector<int> v(n+1),w(n+1);
  rep(i,n){
    cin >> v[i+1] >> w[i+1];
  }
  vector<vector<P>> data((1<<9)+5); //d9 made hanbun-zenrekkyo
  REP(i,1,min(n+1,1<<9)){
    if(i==1){
      data[i].emplace_back(w[i],v[i]);
      data[i].emplace_back(0,0);
    } else {
      data[i] = data[i/2];
      for(auto it:data[i/2]){
        data[i].emplace_back(it.F+w[i],it.S+v[i]);
      }
    }

    sort(data[i].begin(),data[i].end());
    REP(j,1,data[i].size()){
      data[i][j].S = max(data[i][j].S, data[i][j-1].S);
    }
  }

  int Q;
  cin >> Q;
  vector<pair<int,P>> q(Q);
  rep(i,Q){
    cin >> q[i].F >> q[i].S.F;
    q[i].S.S = i;
  }
  sort(q.begin(),q.end());
  vector<int> A(Q);
  unordered_map<int,vector<P>> mp;
  rep(s,Q){
    int x=q[s].F, L=q[s].S.F; 
    int ans;
    if(x<min(n+1,1<<9)){
      int pos = lower_bound(data[x].begin(),data[x].end(),P(L+1,0)) - data[x].begin();
      ans = data[x][pos-1].S;
//      cout << ans << ln;
    } else {
      vector<int> ls;
      int tmp = x;
      while(tmp>=(1<<9) && mp.find(tmp)==mp.end()){ ls.emplace_back(tmp); tmp/=2; }
//      sort(ls.rbegin(),ls.rend());
      int sz = ls.size();
      vector<P> sl;
      rep(b,1<<sz){
        int vsum = 0, wsum = 0;
        rep(j,sz)if(bit(b,j)){
          vsum += v[ls[j]];
          wsum += w[ls[j]];
        }
        if(mp.find(tmp)!=mp.end()){
          for(auto it:mp[tmp]){
            sl.emplace_back(it.F+wsum, it.S+vsum);
          }
        } else {
          sl.emplace_back(wsum,vsum);
        }
      }
      mp[x] = sl;
      while(tmp>=(1<<9)){ tmp/=2; }
      for(auto it:sl){
        if(it.F<=L){
          int pos = lower_bound(data[tmp].begin(),data[tmp].end(),P(L-it.F+1,0)) - data[tmp].begin();
          ans = max(ans, it.S+data[tmp][pos-1].S);
        }
      }
//      cout << ans << ln;
    }
    A[q[s].S.S] = ans;
  }

  for(auto ans:A){
    cout << ans << ln;
  }
  return 0;
}