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)

from collections import defaultdict
from bisect import bisect_right


def f(li, w_max):
    d = defaultdict(int)
    for i in range(1 << len(li)):
        v_sm = 0
        w_sm = 0
        now = i
        cnt = 0
        while now:
            if now & 1:
                v, w = vw[li[cnt]]
                v_sm += v
                w_sm += w

            now >>= 1
            cnt += 1

        if w_sm <= w_max:
            d[w_sm] = max(d[w_sm], v_sm)

    return d


n = int(input())
vw = [(None, None)]
for _ in range(n):
    v, w = map(int, input().split())
    vw.append((v, w))
q = int(input())
ul = [map(int, input().split()) for _ in range(q)]

for u, l in ul:
    li = []
    now = u
    while now:
        li.append(now)
        now //= 2

    ln = len(li)
    d1 = f(li[:ln//2], l)
    d2 = f(li[ln//2:], l)

    wv2 = [[key, val] for key, val in d2.items()]
    wv2.sort()
    for i in range(1, len(wv2)):
        w_now, v_now = wv2[i]
        w_pre, v_pre = wv2[i-1]
        if v_pre > v_now:
            wv2[i][1] = v_pre

    ans = 0
    for k1, v1 in d1.items():
        i = bisect_right(wv2, [l - k1 + 0.5, v1])
        ans = max(ans, wv2[i-1][1] + v1)

    print(ans)