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)

#!usr/bin/env python3
from collections import defaultdict, deque
from heapq import heappush, heappop
from itertools import permutations, accumulate
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.buffer.readline().split()]
def I(): return int(sys.stdin.buffer.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

def solve():
    def dp_all(lis, l):
        res = [(0,0)]
        for v,w in lis:
            res += [(sv+v,sw+w) for (sv,sw) in res if sw+w <= l]
        return max(res)[0]

    def array(lis):
        res = [(0,0)]
        for v,w in lis:
            res += [(sw+w,sv+v) for (sw,sv) in res]
        res.sort(key = lambda x:(x[0],-x[1]))
        sw = []
        sv = []
        maxv = 0
        for i in range(len(res)):
            w,v = res[i]
            if v < maxv:
                continue
            sv.append(v)
            sw.append(w)
            maxv = v
        return sw, sv

    def dp(lis, res_sw, res_sv, l):
        liss = [(0,0)]
        for v,w in lis:
            liss += [(sv+v,sw+w) for (sv,sw) in liss if sw+w <= l]
        ans = 0
        for sv,sw in liss:
            j = bisect.bisect_right(res_sw, l-sw)-1
            s = sv+res_sv[j]
            if ans < s:
                ans = s
        return ans

    n = I()
    g = [None]*(n+1)
    for i in range(1,n+1):
        v,w = LI()
        g[i] = (v,w)
    q = I()
    m = n.bit_length()>>1
    B = 1<<m
    mem_sw = [None]*B
    mem_sv = [None]*B
    for _ in range(q):
        x,l = LI()
        lis = []
        bit = x
        while x:
            lis.append(g[x])
            x >>= 1
        h = len(lis)
        if h < m:
            print(dp_all(lis,l))
        else:
            bit = (bit-(bit&(B-1)))>>m
            if mem_sw[bit] is None:
                mem_sw[bit],mem_sv[bit] = array(lis[-m:])
            print(dp(lis[:-m], mem_sw[bit], mem_sv[bit], l))

    return

#Solve
if __name__ == "__main__":
    solve()