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
sys.setrecursionlimit(10**7) #再帰関数の上限,10**5以上の場合python
import math
from copy import copy, deepcopy
from copy import deepcopy as dcp
from operator import itemgetter
from bisect import bisect_left, bisect, bisect_right#2分探索
#bisect_left(l,x), bisect(l,x)#aはソート済みである必要あり。aの中からx未満の要素数を返す。rightだと以下
from collections import deque
#deque(l), pop(), append(x), popleft(), appendleft(x)
##listでqueの代用をするとO(N)の計算量がかかってしまうので注意
from collections import Counter#文字列を個数カウント辞書に、
#S=Counter(l),S.most_common(x),S.keys(),S.values(),S.items()
from itertools import accumulate,combinations,permutations#累積和
#list(accumulate(l))
from heapq import heapify,heappop,heappush
#heapify(q),heappush(q,a),heappop(q) #q=heapify(q)としないこと、返り値はNone
#import fractions#古いatcoderコンテストの場合GCDなどはここからimportする
from functools import lru_cache#pypyでもうごく
#@lru_cache(maxsize = None)#maxsizeは保存するデータ数の最大値、2**nが最も高効率
import numpy as np
def input(): return sys.stdin.readline()[:-1]
def printl(li): _=print(*li, sep="\n") if li else None
def argsort(s, return_sorted=False): 
    inds=sorted(range(len(s)), key=lambda k: s[k])
    if return_sorted: return inds, [s[i] for i in inds]
    return inds
def alp2num(c,cap=False): return ord(c)-97 if not cap else ord(c)-65
def num2alp(i,cap=False): return chr(i+97) if not cap else chr(i+65)
def matmat(A,B):
    K,N,M=len(B),len(A),len(B[0])
    return [[sum([(A[i][k]*B[k][j]) for k in range(K)]) for j in range(M)] for i in range(N)]
def matvec(M,v):
    N,size=len(v),len(M)
    return [sum([M[i][j]*v[j] for j in range(N)]) for i in range(size)]
def T(M):
    n,m=len(M),len(M[0])
    return [[M[j][i] for j in range(n)] for i in range(m)]
def main():
    mod = 1000000007
    #w.sort(key=itemgetter(1),reversed=True)  #二個目の要素で降順並び替え

    N = int(input())
    #N, K = map(int, input().split())
    #A = tuple(map(int, input().split())) #1行ベクトル
    #L = tuple(int(input()) for i in range(N)) #改行ベクトル
    #S = tuple(tuple(map(int, input().split())) for i in range(N)) #改行行列
    vs=[]
    ws=[]
    costs=tuple(tuple(map(int, input().split())) for i in range(N))
    visited=[0]*N
    Q=int(input())
    qs= tuple(tuple(map(int, input().split())) for i in range(Q))
    maxL=int(max([qs[i][1] for i in range(Q)]))

    
        
    for v,L in qs:
        if v==1:
            v,w=costs[1]
            if w<=L:print(v)
            else:print(0)
            continue

        n=[]
        while v:
            n.append(costs[v-1])
            v=v//2
        ln=len(n)
        n1=n[:ln//2]
        n2=n[ln//2:]
        
        n1c=[]
        n2c=[]
        for i in range(len(n1)+1):
            for n1b in combinations(n1,i):
                vsum=sum([x[0] for x in n1b])
                wsum=sum([x[1] for x in n1b])
                n1c.append((wsum,vsum))
        for i in range(len(n2)+1):
            for n2b in combinations(n2,i):
                vsum=sum([x[0] for x in n2b])
                wsum=sum([x[1] for x in n2b])
                n2c.append((wsum,vsum))
        vmax=10**5+1
        n1c.sort(key=lambda k: k[0]*vmax+k[1])
        n2c.sort(key=lambda k: k[0]*vmax+k[1])
        
        j1=1
        for i in range(1,len(n1c)):
            if n1c[i][1]>n1c[j1-1][1]:
                n1c[j1]=n1c[i]
                j1+=1
        
        j2=1
        for i in range(1,len(n2c)):
            if n2c[i][1]>n2c[j2-1][1]:
                n2c[j2]=n2c[i]
                j2+=1

        j=j1-1
        ans=0
        for i in range(j2):
            w,v=n2c[i]
            while n1c[j][0]+w>L and j>=0:
                j-=1
            if j==-1:
                break
            ans=max(ans,v+n1c[j][1])
        print(ans)







if __name__ == "__main__":
    main()