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 <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <stack>
#include <tuple>
#include <cmath>
#include <iomanip>
#include <map>
#include <cstring> //memset(dp,0,sizeof(dp))
#include <functional>
#include <cctype>
#include <locale>
#define ll long long
#define rep(i,n) for(int i=0;i<(n);i++)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define fi first
#define se second
#define pb push_back
#define ppb pop_back()
#define ALL(a) (a).begin(),(a).end()
#define int long long
using namespace std;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
typedef pair<int,int> P;
using pll=pair<long long,long long>;
//s[i]=tolower(s[i]); islower(s[i]); cout << tolower(s[i])はバグ
pll operator+(pll x,pll y) {
  return make_pair(x.first+y.first,x.second+y.second);
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int nn; cin >> nn;
    vector<int> v1(nn+1),ww(nn+1);
    rep(i,nn) cin >> v1[i+1] >> ww[i+1];
    int q; cin >> q;
    rep(query,q){
        int ve,W; cin >> ve >> W;
        int N=log2(ve)+1;
        int n=N/2;
        vector<pll> w1(n);vector<pll> w2(N-n);
        for (int i=0;i<N;i++) {
            if (i<n){ w1[i].se=v1[ve]; w1[i].fi=ww[ve];}
            else {w2[i-n].se=v1[ve]; w2[i-n].fi=ww[ve];}
            ve/=2;
        }
        vector<pll> W1((1<<n),make_pair(0LL,0LL));vector<pll> W2((1<<(N-n)),make_pair(0LL,0LL));
        for (int bit=0;bit<(1<<n);bit++) {
          for (int i=0;i<n;i++) {
            if ((bit>>i) & 1) W1[bit]=W1[bit]+w1[i];
          }
        }
        for (int bit=0;bit<(1<<(N-n));bit++) {
          for (int i=0;i<(N-n);i++) {
            if ((bit>>i) & 1) W2[bit]=W2[bit]+w2[i];
          }
        }
        sort(W1.begin(),W1.end());vector<pll> WW(0);WW.push_back(W1[0]);
        for (int i=0;i<(1<<n)-1;i++) {
          if (W1[i+1].second>W1[i].second) WW.push_back(W1[i+1]);
        }
        long long max_v=0LL;
        for (int i=0;i<(1<<(N-n));i++) {
          long long wr=W-W2[i].first;
          if (wr<0) continue;
          auto ir=upper_bound(WW.begin(),WW.end(),make_pair(wr,100000000000005LL));ir--;
          pll P=*ir;
          max_v=max(max_v,P.second+W2[i].second);
        }
        cout << max_v << '\n';
    }
}