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)

def examA():
    S = SI()
    ans = S[:3]
    print(ans)
    return

def examB():
    A, V = LI()
    B, W = LI()
    T = I()
    L = abs(A-B)
    v = V-W
    if v*T>=L:
        ans = "YES"
    else:
        ans = "NO"
    print(ans)
    return

def examC():
    N, K = LI()
    A = LI()
    for _ in range(min(60,K)):
        cur = N
        L = [0] * (N + 1)
        R = [0] * (N + 1)
        for i in range(N):
            a = A[i]
            if 0<i-a:
                L[i-a] += 1
            else:
                L[0] += 1
            if N>i+a+1:
                R[i+a+1] += 1
            else:
                R[N] += 1
            if cur>a:
                cur = a
        if cur==N:
            break
        A[0] = L[0]
        for i in range(1,N):
            A[i] = A[i-1]+L[i]-R[i]


    print(" ".join(map(str,A)))
    return

def examD():
    N = I()
    VW = [LI()for _ in range(N)]
    V = [-1]*N
    for i in range(1,N):
        a, b = (i-1)//2, i
        V[b] = a
    half = [[]for _ in range(2048)]

    for i in range(min(N,2048)):
        s = i
        candi = defaultdict(int)
        candi[0] = 0
        for _ in range(12):
            vc, wc = VW[s]
            ne = deepcopy(candi)

            for noww, nowv in candi.items():
                if ne[noww + wc] < nowv + vc:
                    ne[noww + wc] = nowv + vc
            candi = ne
            if s == 0:
                break
            s = V[s]
        half[i] = sorted(candi.items())

    Q = I()
    ans = [0]*Q
    #print(half)
    for q in range(Q):
        s, L = LI()
        s -= 1
        if s<2048:
            cur = 0
            for noww, nowv in half[s]:
                if cur < nowv and noww <= L:
                    cur = nowv
            ans[q] = cur
            continue



        candi = defaultdict(int)
        candi[0] = 0
        for _ in range(10):
            vc, wc = VW[s]
            ne = deepcopy(candi)

            for noww, nowv in candi.items():
                if ne[noww + wc] < nowv + vc:
                    ne[noww + wc] = nowv + vc
            candi = ne
            s = V[s]
            if s<2048:
                break

        #print(candi)
        cur = 0
        for noww, nowv in candi.items():
            if cur<nowv and noww<=L:
                plus = bisect.bisect_left(half[s],L-noww)
                """
                                x = nowv+half[s][min(len(half[s])-1,plus)]
                #x = nowv + half[s][0]
                if cur<x:
                    cur = nowv+x
                """
        ans[q] = cur


    for v in ans:
        print(v)
    return

def examE():
    N, K, S, T = LI()

    ans = 0
    print(ans)
    return

def examF():
    ans = 0
    print(ans)
    return

def test():
    i = I()
    li = LI()
    lsi = LSI()
    si = LS()
    print(i)
    print(li)
    print(lsi)
    print(si)
    return

from decimal import Decimal as dec
import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(input())
def LI(): return list(map(int,sys.stdin.readline().split()))
def DI(): return dec(input())
def LDI(): return list(map(dec,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 10**(-12)
alphabet = [chr(ord('a') + i) for i in range(26)]

sys.setrecursionlimit(10**7)

if __name__ == '__main__':
    examD()

"""
142
12 9 1445 0 1
asd dfg hj o o
aidn
"""