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>
typedef long long LL;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)

using namespace std;

int main(void)
{
  int n;
  cin >> n;
  vector<LL> value(1<<18), weight(1<<18);
  FOR(i,1,n+1) cin >> value[i] >> weight[i];
  vector<vector<pair<LL, LL>>> hanzen(1<<13); // weight, value
  hanzen[0].emplace_back(0,0);
  FOR(bits,1,1<<13) {
    for(pair<LL,LL> p:hanzen[bits/2]) {
      pair<LL,LL> added = make_pair(p.first+weight[bits], p.second+value[bits]);
      hanzen[bits].push_back(p);
      hanzen[bits].push_back(added);
    }
    vector<pair<LL,LL>> tmp(hanzen[bits].begin(),hanzen[bits].end());
    sort(tmp.begin(),tmp.end());
    hanzen[bits].clear();
    hanzen[bits].emplace_back(0,0);
    for(pair<LL,LL> p:tmp) {
      if(hanzen[bits].back().second < p.second) hanzen[bits].push_back(p);
    }
  }
  int q;
  cin >> q;
  while(q--) {
    LL v, l;
    cin >> v >> l;
    vector<pair<LL,LL>> comb;
    comb.emplace_back(0,0);
    while(v>=(1<<13)) {
      vector<pair<LL,LL>> tmp(comb.begin(), comb.end());
      for(pair<LL,LL> p:tmp) {
        pair<LL,LL> added = make_pair(p.first+weight[v], p.second+value[v]);
        comb.push_back(added);
      }
      v/=2;
    }
    LL answer = 0;
    for(pair<LL,LL> p: comb) {
      LL cw,cv;
      tie(cw,cv) = p;
      if(cw>l) continue;
      auto ite = lower_bound(hanzen[v].begin(),hanzen[v].end(), make_pair(l-cw+1,0ll));
      --ite;
      answer=max(answer, cv+ite->second);
    }
    cout << answer << endl;
  }

  return 0;
}