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: Accepted)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		int[] v = new int[n];
		int[] w = new int[n];
		for (int i = 0; i < n; i++) {
			String[] sa = br.readLine().split(" ");
			v[i] = Integer.parseInt(sa[0]);
			w[i] = Integer.parseInt(sa[1]);
		}
		int q = Integer.parseInt(br.readLine());
		int[] x = new int[q];
		int[] l = new int[q];
		for (int i = 0; i < q; i++) {
			String[] sa = br.readLine().split(" ");
			x[i] = Integer.parseInt(sa[0]);
			l[i] = Integer.parseInt(sa[1]);
		}
		br.close();

		int end = Math.min(1 << 11, n + 1);
		int[][] dp = new int[end][100001];
		for (int i = 1; i < end; i++) {
			int i1 = i - 1;
			int pre = i / 2;
			for (int j = 0; j <= 100000; j++) {
				dp[i][j] = dp[pre][j];
			}
			for (int j = 100000 - w[i1]; j >= 0; j--) {
				dp[i][j + w[i1]] = Math.max(dp[i][j + w[i1]], dp[i][j] + v[i1]);
			}
		}

		PrintWriter pw = new PrintWriter(System.out);
		for (int i = 0; i < q; i++) {
			List<Integer> list = new ArrayList<>();
			int now = x[i];
			while (now >= end) {
				list.add(now - 1);
				now /= 2;
			}
			int ans = dp[now][l[i]];
			int size = list.size();
			int end2 = 1 << size;
			for (int j = 0; j < end2; j++) {
				int vs = 0;
				int ws = 0;
				for (int j2 = 0; j2 < size; j2++) {
					if ((j >> j2 & 1) == 1) {
						vs += v[list.get(j2)];
						ws += w[list.get(j2)];
					}
				}
				if (ws <= l[i]) {
					int val = vs + dp[now][l[i] - ws];
					ans = Math.max(ans, val);
				}
			}
			pw.println(ans);
		}
		pw.flush();
	}
}