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 int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define eb emplace_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second

using vint=vector<int>;
using pint=pair<int,int>;
using vpint=vector<pint>;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

template<class A,class B>
ostream& operator<<(ostream& ost,const pair<A,B>&p){
	ost<<"{"<<p.first<<","<<p.second<<"}";
	return ost;
}

template<class T>
ostream& operator<<(ostream& ost,const vector<T>&v){
	ost<<"{";
	for(int i=0;i<v.size();i++){
		if(i)ost<<",";
		ost<<v[i];
	}
	ost<<"}";
	return ost;
}

inline int topbit(unsigned long long x){
	return x?63-__builtin_clzll(x):-1;
}

inline int popcount(unsigned long long x){
	return __builtin_popcountll(x);
}

inline int parity(unsigned long long x){
	return __builtin_parity(x);
}

int N,Q;
int V[1<<19],W[1<<19];

int sumV[1<<9],sumW[1<<9];
void calc(int v,int lim){
	memset(sumV,0,sizeof(sumV));
	memset(sumW,0,sizeof(sumW));
	vint vs,ws;
	while(lim<=topbit(v)){
		vs.pb(V[v]);
		ws.pb(W[v]);
		v>>=1;
	}
	for(int i=1;i<1<<vs.size();i++){
		int k=topbit(i);
		sumV[i]=sumV[i^(1<<k)]+vs[k];
		sumW[i]=sumW[i^(1<<k)]+ws[k];
	}
}


vector<tuple<int,int,int>>bucket[1<<9];

const int X=100000;
int dp[X+1];
int ans[111111];
signed main(){
	scanf("%lld",&N);
	rep(i,N)scanf("%lld%lld",&V[i+1],&W[i+1]);
	scanf("%lld",&Q);
	rep(i,Q){
		int v,L;
		scanf("%lld%lld",&v,&L);
		int w=v;
		w>>=max(0ll,topbit(v)-8);
		bucket[w].eb(v,L,i);
	}

	for(int i=1;i<=N&&i<(1<<9);i++){
		calc(i,0);
		memset(dp,0,sizeof(dp));
		rep(j,1<<9){
			if(sumW[j]>X)continue;
			chmax(dp[sumW[j]],sumV[j]);
		}
		for(int j=0;j<X;j++)chmax(dp[j+1],dp[j]);

		for(auto &tpl:bucket[i]){
			int v,L,k;
			tie(v,L,k)=tpl;

			calc(v,9);
			rep(j,1<<9){
				if(sumW[j]>L)continue;
				chmax(ans[k],dp[L-sumW[j]]+sumV[j]);
			}
		}
	}

	rep(i,Q)printf("%lld\n",ans[i]);
	return 0;
}