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: Runtime Error)

import sys
from heapq import heappop, heappush

readline = sys.stdin.readline
readall = sys.stdin.read
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prn = lambda x: print(*x, sep='\n')


def solve():
    n = ni()
    bag = [tuple(nm()) for _ in range(n)]
    q = ni()
    query = [tuple(nm()) for _ in range(q)]
    ansd = dict()
    g = [None]* 2 ** 10

    def calc(v):
        W = 10**5 + 1
        res = [0] * W
        l = list()
        while v:
            l.append(bag[v-1])
            v >>= 1
        m = len(l)
        for bit in range(1 << m):
            cv = cw = 0
            for i in range(m):
                if bit & (1 << i):
                    cv += l[i][0]
                    cw += l[i][1]
                    if cw > W:
                        break
            else:
                if res[cw] < cv:
                    res[cw] = cv
        for i in range(W-1):
            if res[i+1] < res[i]:
                res[i+1] < res[i]
        return res


    for _v, L in query:
        v = _v
        if v < 2 ** 10:
            l = list()
            while v:
                l.append(bag[v-1])
                v >>= 1
            m = len(l)
            ans = 0
            for bit in range(1 << m):
                cv = cw = 0
                for i in range(m):
                    if bit & (1 << i):
                        cv += l[i][0]
                        cw += l[i][1]
                        if cw > L:
                            break
                else:
                    if cv > ans:
                        ans = cv

            ansd[(_v, L)] = ans

        else:
            l = list()
            while v >= 2 ** 10:
                l.append(bag[v-1])
                v >>= 1
            if g[v] is None:
                g[v] = calc(v)
            m = len(l)
            ans = 0
            for bit in range(1 << m):
                cv = cw = 0
                for i in range(m):
                    if bit & (1 << i):
                        cv += l[i][0]
                        cw += l[i][1]
                        if cw > L:
                            break
                else:
                    if cv + g[v][L-cw] > ans:
                        ans = cv + g[v][L-cw]
            ansd[(_v, L)] = ans

    for c in query:
        print(ansd[c])
    return

solve()