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)

import sys
input = sys.stdin.readline
n = int(input())
vw = [[0,0]]+[list(map(int,input().split())) for i in range(n)]
q = int(input())
query = [list(map(int,input().split())) for i in range(q)]
border = 12
knapls = [[] for i in range(2**border)]
knapls[0].append((0,0))
for i in range(1,2**border):
  ls = []
  if i > n:
    break
  ls.extend(knapls[i//2])
  v,w = vw[i]
  l = len(ls)
  for j in range(l):
    vp,wp = ls[j]
    if wp+w > 10**5:
      continue
    ls.append((vp+v,wp+w))
  ls.sort(key = lambda x:x[1])
  knapls[i].append((0,0))
  for j in range(len(ls)):
    vn,wn = ls[j]
    vp,wp = knapls[i][-1]
    if wp <= wn and vp >= vn:
      continue
    knapls[i].append(ls[j])
for v,L in query:
  if v >= 2**border:
    anc = v
    ls = [(0,0)]
    while anc >= 2**border:
      vn,wn = vw[anc]
      l = len(ls)
      for i in range(l):
        vp,wp = ls[i]
        if wp+wn > L:
          continue
        ls.append((vp+vn,wp+wn))
      anc //= 2
    ans = 0
    l = len(ls)
    lnap = len(knapls[anc])
    for i in range(l):
      vn,wn = ls[i]
      rest = L-wn
      left = 0
      right = lnap
      if i and wn >= ls[i-1][1] and vn <= ls[i-1][0]:
        continue
      while left+1 < right:
        mid = (left+right)//2
        if knapls[anc][mid][1] > rest:
          right = mid
        else:
          left = mid
      ans = max(knapls[anc][left][0]+vn,ans)
    print(ans)
  else:
    ans = 0
    left = 0
    right = len(knapls[v])
    while left+1 < right:
      mid = (left+right)//2
      if knapls[v][mid][1] > L:
        right = mid
      else:
        left = mid
    ans = knapls[v][left][0]
    print(ans)