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.*;

public class Main {

    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int MAX = 100000;
        int[][] map = new int[512][MAX+1];

        int[] v = new int[n];
        int[] w = new int[n];
        for(int i=0; i<n; i++) {
            v[i] = sc.nextInt();
            w[i] = sc.nextInt();
        }

        int pn = 9;

        for(int k=w[0]; k<=MAX; k++) {
            map[1][k] = v[0];
        }
        for(int i=2; i<(1<<pn); i++) {
            int j = i/2;
            if(i > n) break;
            for(int k = 0; k <= MAX; k++) {
                map[i][k] = Math.max(map[i][k], map[j][k]);
                if(k + w[i-1] > MAX) continue;
                map[i][k+w[i-1]] = Math.max(map[i][k+w[i-1]], map[j][k]+v[i-1]);
            }
        }

        int q = sc.nextInt();

        for(int i=0; i<q; i++) {
            int x = sc.nextInt();
            int l = sc.nextInt();
            if(x < (1<<pn)) {
                System.out.println(map[x][l]);
                continue;
            }
            int xl = 32 - Integer.numberOfLeadingZeros(x);
            int m = xl-pn;
            int[] path = new int[m];
            for(int j=0; j<m; j++) {
                path[j] = x >> j;
            }
            int base = x >> m;
            int max = 0;
            for(int j=0; j<1<<m; j++) {
                int cost = 0;
                int value = 0;
                for(int k=0; k<m; k++) {
                    if((j>>k)%2 == 1) {
                        cost += w[path[k]-1];
                        value += v[path[k]-1];
                    }
                }
                if(cost <= l) {
                    max = Math.max(max, map[base][l-cost] + value);
                }
            }
            System.out.println(max);
        }
    }




}