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)

#from __future__ import division

import sys
class Knapsack:
    def __init__(self, VW):
        self.VW = VW
        self.VW.sort(key=lambda vw: vw[0] / vw[1], reverse=True)
        self.n = len(VW)
 
    def solve(self, capacity):
        self.capacity = capacity
        self.ok = self._greedy()
        return self._dfs(0, 0, 0)
 
    def _greedy(self):
        capacity = self.capacity
        res = 0
        for v, w in self.VW:
            if w <= capacity:
                res += v
                capacity -= w
        return res
 
    def _dfs(self, depth, s0, s1):
        ng, f = self._solve_relaxation(s0, s1)
        if f == -1:
            self.ok = max(self.ok, ng)
            return ng
        if ng < self.ok:
            return -float("inf")
        res = -float("inf")
        res = max(res, self._dfs(depth+1, s0 | 1<<f, s1))
        res = max(res, self._dfs(depth+1, s0, s1 | 1<<f))
        return res
 
    def _solve_relaxation(self, s0, s1):
        f = -1
        res = 0
        capacity = self.capacity
        for i, (v, w) in enumerate(self.VW):
            if s1 >> i & 1:
                res += v
                capacity -= w
        if capacity < 0:
            return -float("inf"), f
        forbidden = s0 | s1
        for i, (v, w) in enumerate(self.VW):
            if forbidden >> i & 1:
                continue
            if capacity == 0:
                break
            if w <= capacity:
                capacity -= w
                res += v
            else:
                f = i
                res += v * (capacity / w)
                break
        return res, f

def main():
    input = sys.stdin.readline
    N = int(input())
    VW = [0] + [list(map(int, input().split())) for _ in range(N)]
    Q = int(input())
    Ans = []
    for _ in range(Q):
        V, L = map(int, input().split())
        vw = []
        while V > 0:
            vw.append(VW[V])
            V //= 2
        knapsack = Knapsack(vw)
        ans = knapsack.solve(L)
        Ans.append(ans)
    print("\n".join(map(str, Ans)))

main()