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# (Mono-mcs 6.8.0.105) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
class Program
{
	static void Main(string[] args)
	{
		int N = int.Parse(Console.ReadLine());
		int[]V = new int[N];
		int[]W = new int[N];
		for(var i=0;i<N;i++){
			string[] str = Console.ReadLine().Split();
			V[i] = int.Parse(str[0]);
			W[i] = int.Parse(str[1]);
		}
		int Q = int.Parse(Console.ReadLine());
		var list = new List<int>[Q];
		for(var i=0;i<Q;i++){
			string[] str = Console.ReadLine().Split();
			list[i] = new List<int>();
			list[i].Add(i);
			list[i].Add(int.Parse(str[0]));
			list[i].Add(int.Parse(str[1]));
			int a1 = list[i][1];
			while(a1<(int)Math.Pow(2,N-1)){
				a1 = a1 * 2 + 1;
			}
			list[i].Add(a1);
		}
		Array.Sort(list,(x,y)=>x[3]-y[3]);
		int[]H = new int[20];
		var ans = new List<int>[Q];
		int[,]D = new int[20,100001];
		for(var i=0;i<Q;i++){
			var listA = new List<int>();
			var ct = list[i][1];
			listA.Add(list[i][1]);
			while(ct>1){
				ct /= 2;
				listA.Add(ct);
			}
			for(var j=1;j<=listA.Count;j++){
				var ca = listA[listA.Count-j];
				if(H[j-1] != ca){
					for(var k=1;k<=100000;k++){
						if(k - W[ca-1] >= 0){
							D[j,k] = Math.Max(D[j-1,k],D[j-1,k-W[ca-1]]+V[ca-1]);
						} else {
							D[j,k] = D[j-1,k];
						}
						D[j,k] = Math.Max(D[j,k-1],D[j,k]);
					}
				}
				H[j-1] = ca;
			}
			ans[i] = new List<int>{list[i][0],D[listA.Count,list[i][2]]};
			for(var j=listA.Count+1;j<20;j++){
				H[j-1] = 0;
			}
		}
		//Array.Sort(ans,(x,y)=>x[0]-y[0]);
		for(var i=0;i<Q;i++){
			Console.WriteLine(ans[i][1]);
		}
	}
}