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: Time Limit Exceeded)

import sys
import numpy as np
from numba import njit
input = sys.stdin.buffer.readline

N = int(input())
vws = [tuple(map(int, input().split())) for _ in range(N)]

n = min(2**10, N)
vws_arr = np.array(vws)

dp = np.zeros((n, 10**5+1), np.int32)
dp[0, vws[0][1]:] = vws[0][0]
for i in range(1, n):
    v, w = vws[i]
    dp[i] = dp[(i-1)//2]
    dp[i, w:] = np.maximum(dp[(i-1)//2, :-w]+v, dp[i, w:])


@njit("i4(i4[:], i4[:], i4, i4[:])", cache = True)
def func(vs, ws, L, dp_):
    bits = np.arange(1 << len(vs))
    values = np.zeros_like(bits, np.int32)
    weights = np.zeros_like(bits, np.int32)
    for i in range(len(vs)):
        indices = np.nonzero(np.bitwise_and(bits, 1<<i))
        values[indices] += vs[i]
        weights[indices] += ws[i]

    mask = (weights <= L)     
    return np.amax(dp_[L - weights[mask]] + values[mask])

Q = int(input())
for _ in range(Q):
    V, L = map(int, input().split())
    V -= 1
    if V < n:
        print(dp[V, L])
    else:
        Vs = []
        while V >= n:
            Vs.append(V)
            V = (V-1) // 2
        vs = vws_arr[Vs, 0]
        ws = vws_arr[Vs, 1]
        print(func(np.array(vs, np.int32), np.array(ws, np.int32), L, dp[V]))