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
input = lambda: sys.stdin.readline().rstrip()

N = int(input())
M = 10 ** 5 + 1
K = 9
A = [(0, 0)] * max(N + 1, 1 << K)
X = [[], [-1] * M]
X[1][0] = 0
v, w = map(int, input().split())
X[1][w] = v
A[1] = (v, w)
for i in range(2, min(N + 1, 1 << K)):
    v, w = map(int, input().split())
    if i % 2 == 0:
        X.append(X[i // 2][:])
    else:
        X.append([])
        X[i] = X[i // 2]
    
    for j in range(w, M)[::-1]:
        if X[i][j-w] >= 0:
            X[i][j] = max(X[i][j], X[i][j-w] + v)
    
    A[i] = (v, w)

for i in range(1 << K, N + 1):
    v, w = map(int, input().split())
    A[i] = (v, w)

if len(X) < 1 << K:
    for i in range(len(X), 1 << K):
        X.append([-1] * M)

for i in range(1 << K - 1, 1 << K):
    for j in range(1, M):
        X[i][j] = max(X[i][j], X[i][j-1])

Q = int(input())
for _ in range(Q):
    i, maw = map(int, input().split())
    if i < 1 << K - 1:
        V = []
        W = []
        while i:
            v, w = A[i]
            V.append(v)
            W.append(w)
            i //= 2
        k = len(V)
        ma = 0
        WW = [0] * (1 << k)
        VV = [0] * (1 << k)
        for i in range(1, 1 << k):
            a = i & (-i)
            b = a.bit_length() - 1
            ww = WW[i-a] + W[b]
            WW[i] = ww
            vv = VV[i-a] + V[b]
            VV[i] = vv
            if ww <= maw:
                ma = max(ma, vv)
        print(ma)
    else:
        V = []
        W = []
        while i >= 1 << K:
            v, w = A[i]
            V.append(v)
            W.append(w)
            i //= 2
        ii = i
        k = len(V)
        WW = [0] * (1 << k)
        VV = [0] * (1 << k)
        ma = X[ii][maw]
        for i in range(1, 1 << k):
            a = i & (-i)
            b = a.bit_length() - 1
            ww = WW[i-a] + W[b]
            WW[i] = ww
            vv = VV[i-a] + V[b]
            VV[i] = vv
            if ww > maw: continue
            ma = max(ma, X[ii][maw - ww] + vv)
        print(ma)