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)

#!/usr/bin/env python3
import sys

#solve
def solve():
    input = sys.stdin.buffer.readline
    LI = lambda: map(int, input().split())
    II = lambda: int(input())
    n = II()
    vw = [[]] * n
    nbit = 1 << n.bit_length()
    M = 10 ** 5 + 1
    M0 = [0] * M
    XX = 1 << n.bit_length() // 2
    d = [[]] * XX
    d[0] = M0
    for i in range(1, XX):
        v, w = LI()
        vw[i - 1] = (v, w)
        di = d[i // 2][:]
        for j in range(M - 1, w - 1, -1):
            if di[j] < di[j - w] + v:
                di[j] = di[j - w] + v
        d[i] = di
    for i in range(XX, n + 1):
        v, w = LI()
        vw[i - 1] = (v, w)
    n = [pow(2, i) for i in range(18)]
    for _ in range(II()):
        V, L = LI()
        lent = V.bit_length()
        if V < XX:
            print(d[V][L])
            continue
        n1 = lent // 2
        n2 = lent - n1
        vw2 = [[]] * n2
        for i in range(n2):
            vw2[i] = (vw[V - 1][0], vw[V - 1][1])
            V //= 2
        lw = [0] * n[n2]
        lv = [0] * n[n2]
        dv = d[V]
        ans = dv[L]
        for mask in range(1, n[n2]):
            a = mask & -mask
            b = a.bit_length() - 1
            v, w = vw2[b]
            maska = mask - a
            ww = lw[maska] + w
            Lww = L - ww
            lw[mask] = ww
            if Lww < 0: continue
            vv = lv[maska] + v
            lv[mask] = vv
            tmp = dv[Lww] + vv
            if tmp > ans:
                ans = tmp
        print(ans)
    return

if __name__ == "__main__":
    solve()