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>

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define FOR(i,n,m) for(int i=(int)(n); i<=(int)(m); i++)
#define RFOR(i,n,m) for(int i=(int)(n); i>=(int)(m); i--)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define RITR(x,c) for(__typeof(c.rbegin()) x=c.rbegin();x!=c.rend();x++)
#define setp(n) fixed << setprecision(n)

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

#define ll long long
#define vll vector<ll>
#define vi vector<int>
#define pll pair<ll,ll>
#define pi pair<int,int>

#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define ins insert

using namespace std;

const int INF = 1e9;
const int MAX_K = 1<<10;

struct P{
    int v,w;
    P(){}
    P(int v, int w):v(v),w(w){}
};
bool operator<(const P &a, const P &b){return a.w<b.w;}
P operator+(const P &a, const P &b){return {a.v+b.v,a.w+b.w};}
ostream &operator<<(ostream &os, const P &a){
    return os<<"("<<a.w<<","<<a.v<<")";
}

vector<P> S[MAX_K];

//-------------------------------------------------

int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N; cin>>N;
    vector<P> nodes(N);
    rep(i,N){
        int v,w; cin>>v>>w;
        nodes[i] = P(v,w);
    }
    S[0].pb(P(0,0));
    int K = min(MAX_K,N+1);
    FOR(i,1,K-1){
        int p = i/2;
        for(P e:S[p]){
            S[i].pb(e);
            S[i].pb(e+nodes[i-1]);
        }
        sort(all(S[i]));
        int M = S[i].size();
        int m=1;
        FOR(j,1,M-1){
            if (S[i][m-1].v < S[i][j].v){
                S[i][m++] = S[i][j];
            }
        }
        S[i].erase(S[i].begin()+m, S[i].end());
    }
    // for(P e:nodes){
    //     cout<<e<<" ";
    // }cout<<endl;
    // rep(v,K){
    //     for(P e:S[v]){
    //         cout<<e<<" ";
    //     }cout<<endl;
    // }

    int Q; cin>>Q;
    while(Q--){
        int v,L; cin>>v>>L;
        vector<P> ps;
        ps.pb(P(0,0));
        while(v>=K){
            int M = ps.size();
            rep(i,M){
                ps.pb(ps[i]+nodes[v-1]);
            }
            v>>=1;
        }
        int ans=0;
        for(P e:ps){
            if (L<e.w) continue;
            auto itr = upper_bound(all(S[v]), P(0,L-e.w)); itr--;
            chmax(ans,e.v+itr->v);
        }
        cout<<ans<<"\n";
    }
    return 0;
}