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: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
class d{
  public static void Main(){
    long n = long.Parse(Console.ReadLine());
    Node[] tree = new Node[n+1];
    long maxw = 0;
    tree[0] = new Node(0,0);
    for(long i=1; i<=n; i++){
      String[] str = Console.ReadLine().Split(' ');
      tree[i] = new Node(long.Parse(str[0]), long.Parse(str[1]));
      if(maxw < long.Parse(str[1])){
        maxw = long.Parse(str[1]);
      }
    }
    long q = long.Parse(Console.ReadLine());
    for(long x=0; x<q; x++){
      String[] order = Console.ReadLine().Split(' ');
      long size = (long)Math.Max(maxw, long.Parse(order[1]));
      long[,] dp = new long[21, size+1];
      Console.WriteLine(knapsack(long.Parse(order[0]), long.Parse(order[1]), dp, tree, 20));
    }
  }
  public static long knapsack(long v, long w, long[,] dp, Node[] t, long d){
    //Console.WriteLine("{0} {1}",v,w);
    if(dp[d,w] != 0){
      if(dp[d,w] == -1){
        return 0;
      }
      else{
        return dp[d,w];
      }
    }
    if(v == 0){
      dp[d,w] = -1;
      return 0;
    }
    if(t[v].W > w){
      long t0 = knapsack(v/2, w, dp, t, d-1);
      dp[d,w] = t0;
      return t0;
    }
    else{
      long t1 = knapsack(v/2, w, dp, t, d-1);
      long t2 = knapsack(v/2, w-t[v].W, dp, t, d-1) + t[v].V;
      if(t1 > t2){
        dp[d,w] = t1;
        return t1;
      }
      else{
        dp[d,w] = t2;
        return t2;
      }
    }
  }
}
class Node{
  public long V{ get; set; }
  public long W{ get; set; }
  public Node(long vt, long wt){
    V = vt;
    W = wt;
  }
}