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.Linq;

class Program
{
    static int X = 9;
    static int XN = 1<<X; // = 512
    static int LMAX = 100000;
    static int[,] dp = new int[XN, LMAX+1];
    
    static int[,] VW;

    static int Dfs(int i, int L)
    {
        if(L<0 ) return 0;
        if(i<XN) return dp[i,L];

        var par = i/2;
        var v = VW[i,0];
        var w = VW[i,1];
        return Math.Max(Dfs(par, L), Dfs(par, L-w) + v);
    }
    
    static void Main()
    {
        var N = int.Parse(Console.ReadLine());
        VW = new int[N+1,2];
        for(int i=1; i<=N; ++i) {
            var vw = Console.ReadLine().Split().Select(int.Parse).ToArray();
            var v = VW[i,0] = vw[0];
            var w = VW[i,1] = vw[1];

            if(i<XN) {
                for(int wi=0; wi<=LMAX; ++wi) {
                    dp[i,wi] = dp[i/2,wi];
                    if(wi-w>=0) dp[i,wi] = Math.Max(dp[i,wi], dp[i/2, wi-w]+v);
                }
            }
        }

        /*
        for(int i=0; i<10; ++i) {
            Console.Write($"{i} : ");
            for(int j=0; j<20; ++j) Console.Write($"{dp[i,j]} ");
            Console.WriteLine();
        }*/
        
        var Q = int.Parse(Console.ReadLine());
        for(int iq=0; iq<Q; ++iq) {
            var vL = Console.ReadLine().Split().Select(int.Parse).ToArray();
            int v = vL[0], L = vL[1];

            Console.WriteLine(Dfs(v, L));
        }

    }
}