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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Main {
	static class Pair implements Comparable<Pair>{
		int weight;
		int val;
		Pair(int w , int v){
			weight = w;
			val = v;
		}
		@Override
		public int compareTo(Pair o) {
			if(weight == o.weight){
				return o.val - val;
			}
			return weight - o.weight;
		}
		@Override
		public String toString() {
			return weight + " " + val;
		}
	}
	static Map<Integer, List<Pair>> memo;
	static List<Pair> memo(int v , int L , int[] V , int W[]){
		if(memo.containsKey(v)){
			return memo.get(v);
		}
		int memoV = v;
		int N = Integer.toBinaryString(v).length();
		int vs[] = new int[N];
		int ws[] = new int[N];
		int index = 0;
		while(v > 0){
			vs[index] = V[v - 1];
			ws[index] = W[v - 1];
			index++;
			v = v / 2;
		}
		List<Pair> ps = new ArrayList<Pair>();
		int result = 0;
		for(int i = 0 ; i < (1 << N) ; ++i){
			int val = 0;
			int weight = 0;
			for(int j = 0 ; j < N ; ++j){
				if(((1 << j) & i) != 0){
					val += vs[j];
					weight += ws[j];
				}
			}
			result = Math.max(result, val);
			ps.add(new Pair(weight , val));
		}
		Collections.sort(ps);
		int mv = 0;
		for(Pair p : ps){
			p.val = Math.max(p.val, mv);
			mv = Math.max(p.val, mv);
		}
		memo.put(memoV, ps);
		return ps;
	}
	static int solve(int v , int L , int[] V , int W[]){
		int B = 12;
		if(v < (1 << B)){
			List<Pair> ps = memo(v , L , V , W);
			int left = 0;
			int right = ps.size() - 1;
			if(ps.get(right).weight <= L){
				return ps.get(right).val;
			}
			while(right - left > 1){
				int mid = (left + right) / 2;
				Pair p = ps.get(mid);
				if(p.weight <= L){
					left = mid;
				}else{
					right = mid;
				}
			}
			return ps.get(left).val;
		}
		int N = Integer.toBinaryString(v).length() - B;
		int vs[] = new int[N];
		int ws[] = new int[N];
		int index = 0;
		while(v >= (1 << B)){
			vs[index] = V[v - 1];
			ws[index] = W[v - 1];
			index++;
			v = v / 2;
		}
		int result = 0;
		List<Pair> ps = memo(v , L , V , W);
		for(int i = 0 ; i < (1 << N) ; ++i){
			int val = 0;
			int weight = 0;
			for(int j = 0 ; j < N ; ++j){
				if(((1 << j) & i) != 0){
					val += vs[j];
					weight += ws[j];
				}
			}
			if(weight > L){
				continue;
			}
			int leftWeight = L - weight;
			int left = 0;
			int right = ps.size() - 1;
			if(ps.get(right).weight <= leftWeight){
				result = Math.max(result, val + ps.get(right).val);				
				continue;
			}
			while(right - left > 1){
				int mid = (left + right) / 2;
				Pair p = ps.get(mid);
				if(p.weight <= leftWeight){
					left = mid;
				}else{
					right = mid;
				}
			}
			result = Math.max(result, val + ps.get(left).val);
		}
		return result;
	}
	public static void main(String[] args) throws Exception{
		memo = new HashMap<Integer, List<Pair>>();
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String line = br.readLine();
		int N = Integer.parseInt(line);
		int V[] = new int[N];
		int W[] = new int[N];
		for(int i = 0 ; i < N ; ++i){
			line = br.readLine();
			String a[] = line.split(" ");
			V[i] = Integer.parseInt(a[0]);
			W[i] = Integer.parseInt(a[1]);
		}
		line = br.readLine();
		int Q = Integer.parseInt(line);
		for(int i = 0 ; i < Q ; ++i){
			line = br.readLine();
			String a[] = line.split(" ");
			int v = Integer.parseInt(a[0]);
			int L = Integer.parseInt(a[1]);
			System.out.println(solve(v, L, V, W));
		}
	}
}