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 Python (3.8.2) (judge verdict: Runtime Error)

import sys
import numpy as np

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines


def main(N, V, W, v, L):
    def precompute(N, V, W):
        U = 10**5 + 10
        dp = np.zeros((1024, U), np.int32)
        for v in range(1, min(1024, N + 1)):
            p = v // 2
            dp[v] = dp[p]
            dp[v, W[v]:] = np.maximum(dp[v, W[v]:], dp[p, :-W[v]] + V[v])
            for i in range(1, U):
                dp[v][i] = max(dp[v][i], dp[v][i - 1])
        return dp

    dp = precompute(N, V, W)
    values = np.empty(1024, np.int32)
    weights = np.empty(1024, np.int32)

    def solve(i, lim, values, weights, dp):
        if i < 1024:
            return dp[i][lim]
        values[0] = 0
        weights[0] = 0
        p = 0
        while i >= 1024:
            for j in range(1 << p):
                values[j + (1 << p)] = values[j] + V[i]
                weights[j + (1 << p)] = weights[j] + W[i]
            p += 1
            i >>= 1
        best = 0
        for n in range(1 << p):
            if weights[n] > lim:
                continue
            x = dp[i][lim - weights[n]] + values[n]
            best = max(best, x)
        return best

    Q = len(v)
    for i in range(Q):
        x = solve(v[i], L[i], values, weights, dp)
        print(x)

        
if sys.argv[-1] == 'ONLINE_JUDGE':
    from numba.pycc import CC
    cc = CC('my_module')
    cc.export('main', '(i4,i4[::1],i4[::1],i4[::1],i4[::1])')(main)
    cc.compile()
    exit()

from my_module import main

N = int(readline())
stdin = np.array(read().split(), np.int32)
V = np.zeros(N + 1, np.int32)
W = np.zeros(N + 1, np.int32)
V[1:] = stdin[:N + N:2]
W[1:] = stdin[1:N + N:2]
query = stdin[N + N + 1:]
v, L = query[::2], query[1::2]

main(N, V, W, v, L)