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 ll long long int
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds;   
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update> 
#define ld long double
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define all(a) a.begin(),a.end()
#define sz(a) (ll)(a.size())
const int M=3e5+5;
int val[M],wt[M];
pair<int,int> res[2][M];
vector<pair<int,int>> vec[M];
int main()
{
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n;
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		cin>>val[i]>>wt[i];
	}
	vec[0].pb({0,0});
	for(int i=1;i<(1<<10);i++)
	{
		int p=i/2;
		for(int j=0;j<sz(vec[p]);j++)
		{
			vec[i].pb(vec[p][j]);
			vec[i].pb({vec[p][j].ff+wt[i],vec[p][j].ss+val[i]});
		}
		sort(all(vec[i]));
		for(int j=1;j<sz(vec[i]);j++)
			vec[i][j].ss=max(vec[i][j].ss,vec[i][j-1].ss);
	}
	int q;
	cin>>q;
	while(q--)
	{
		int v,L;
		cin>>v>>L;
		vector<int> vt;
		while(v>=(1<<10))
		{
			vt.pb(v);
			v/=2;
		}
		vector<pair<int,int>> res;
		for(int i=0;i<(1<<sz(vt));i++)
		{
			int x=0,y=0;
			for(int j=0;j<sz(vt);j++)
			{
				if((i&(1<<j))==0)
					continue;
				x+=val[vt[j]];
				y+=wt[vt[j]];
			}
			res.pb({y,x});
		}
		sort(all(res));
		int curr=sz(vec[v])-1;
		int ans=0;
		for(int i=0;i<sz(res);i++)
		{
			while(curr>=0)
			{
				if(res[i].ff+vec[v][curr].ff>L)
					curr--;
				else
					break;
			}
			if(curr<0)
				break;
			ans=max(ans,res[i].ss+vec[v][curr].ss);
		}
		cout<<ans<<"\n";
	}
	return 0;
}