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: Time Limit Exceeded)

#pragma GCC optimize("Ofast")

#include<cstdio>
#include<vector>
#include<algorithm>

using namespace std;

struct item
{
	int v;long long w;item(int vv=0,long long ww=0):v(vv),w(ww){}
}x[1<<18];
bool operator<(item a,item b){return a.v<b.v;}
item operator+(item a,item b){return item(a.v+b.v,a.w+b.w);}
int n=0;

item tmp[20];int cnt;
item A[1<<10],B[1<<10];int cntA,cntB;

int lowbit(int x){return x&-x;}
int lg[1<<10];

void cons(item* x,int n,int l,item* ans)
{
	for(int i=1;i<(1<<n);i++)
	{
		ans[i]=ans[i-lowbit(i)]+x[lg[lowbit(i)]];
		if(ans[i].v>l)ans[i].v=l+1;
	}
}

int main()
{
	scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d%lld",&x[i].w,&x[i].v);
	for(int i=1;i<1024;i++)lg[i]=lg[i/2]+1;
	int q=0;scanf("%d",&q);
	while(q--)
	{
		int pos=0,l=0;scanf("%d%d",&pos,&l);
		cnt=0;
		while(pos)
		{
			tmp[++cnt]=x[pos];pos=pos/2;
		}
		int mid=cnt/2;
		cons(tmp,mid,l,A),cons(tmp+mid,cnt-mid,l,B);cntA=1<<mid,cntB=1<<(cnt-mid);
		sort(A,A+cntA),sort(B,B+cntB);
		
		//for(int i=0;i<A.size();i++)printf("%d %d\n",A[i].v,A[i].w);
		//puts("");
		//for(int i=0;i<B.size();i++)printf("%d %d\n",B[i].v,B[i].w);
		long long ans=0,maxn=0;
		for(int i=cntA-1,j=-1;i>=0;i--)
		{
			while(j+1<cntB&&A[i].v+B[j+1].v<=l){maxn=max(maxn,B[j+1].w);j++;}
			if(j!=-1)ans=max(ans,A[i].w+maxn);
		}
		printf("%lld\n",ans);
	}
	return 0;
}