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 PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

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

n = int(readline())
data = list(map(int,read().split()))

V = data[:2*n:2]
W = data[1:2*n:2]
q = data[2*n]
v = data[2*n+1::2]
L = data[2*n+2::2]

if(n==1):
    ans = []
    for i,j in zip(v,L):
        if(W[i-1] <= j):
            ans.append(V[0])
        else:
            ans.append(0)
    print('\n'.join(map(str,ans)))
    exit()

# 前処理　最大2**9まで前計算
b_len = min(8, n.bit_length()-1)
bef = [ [0]*(10**5 + 1) for _ in range(2**b_len)]

for i in range(1,2**b_len):
    iw,iv = W[i-1],V[i-1]
    par = i//2
    for j in range(1,10**5+1):
        bef[i][j] = bef[par][j]
        if(j>=iw):
            bef[i][j] = max(bef[i][j] , bef[par][j-iw]+iv)

ans = []
for vi,Li in zip(v,L):
    if(vi==1):
        if(W[0] <= Li):
            ans.append(V[0])
        else:
            ans.append(0)
        continue

    aft = [(0,0)]
    x = vi
    while(x >= 2**b_len):
        xw,xv = W[x-1],V[x-1]
        aft2 = aft[::]
        for jw,jv in aft:
            aft2.append((xw+jw,xv+jv))
        x //= 2
        aft = aft2[::]

    max_ans = 0
    for aw,av in aft:
        lim_w = Li - aw
        if(lim_w < 0):
            continue
        max_ans = max(max_ans, bef[x][lim_w]+av)

    ans.append(max_ans)

print('\n'.join(map(str,ans)))