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.util.*;
import java.io.*;

public class Main {
    static int[] values;
    static int[] weights;
    static int[] maxes;
    static int n;
    static ArrayList<HashMap<Integer, Integer>> dp = new ArrayList<>();
	public static void main (String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		n = Integer.parseInt(br.readLine());
		values = new int[n + 1];
		weights = new int[n + 1];
		maxes = new int[n + 1];
		dp.add(new HashMap<>());
		for (int i = 1; i <= n; i++) {
		    String[] line = br.readLine().split(" ", 2);
		    values[i] = Integer.parseInt(line[0]);
		    weights[i] = Integer.parseInt(line[1]);
    		dp.add(new HashMap<>());
		}
		setMax(1, 0);
		int q = Integer.parseInt(br.readLine());
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < q; i++) {
		    String[] line = br.readLine().split(" ", 2);
		    sb.append(dfw(Integer.parseInt(line[0]), Integer.parseInt(line[1]))).append("\n");
		}
		System.out.print(sb);
	}
	
	static void setMax(int idx, int value) {
	    if (idx > n) {
	        return;
	    }
	    maxes[idx] += value + weights[idx];
	    setMax(idx * 2, maxes[idx]);
	    setMax(idx * 2 + 1, maxes[idx]);
	}
	
	static int dfw(int idx, int value) {
	    if (idx <= 0) {
	        return 0;
	    }
        if (dp.get(idx).containsKey(value)) {
            return dp.get(idx).get(value);
	    }
	    int max = 0;
	    if (value < weights[idx]) {
	        max = dfw(idx / 2, value);
	    } else if (maxes[idx] <= value) {
	        max = dfw(idx / 2, value - weights[idx]) + values[idx];
	    } else {
	        max = Math.max(dfw(idx / 2, value), dfw(idx / 2, value - weights[idx]) + values[idx]);
	    }
	    dp.get(idx).put(value, max);
	    return max;
	}
}