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)

N = int(input())
VW = tuple(tuple(map(int, input().split())) for _ in range(N))
q = int(input())
def f(x, W):
    S = []
    b = x.bit_length()
    for j in range(b):
        S.append(VW[(x>>j)-1])
    def make(S):
        T = {0: 0}
        for v, w in S:
            # 価値v, 重さw
            T0 = dict(T)
            for k, val in T.items():
                if k+w > W:
                    continue
                if k+w in T0:
                    T0[k+w] = max(T0[k+w], val + v)
                else:
                    T0[k+w] = val + v
            T = T0
        v = 0
        R = []
        for k in sorted(T):
            v = max(v, T[k])
            # (重さk, 重さk以下での価値の最大v)
            R.append((k, v))
        return R

    def solve(T0, T1):
        T0.sort()
        T1.sort(reverse=1)

        it = iter(T1); k1, v1 = next(it)
        yield 0
        for k0, v0 in T0:
            # 重さの和がW以下になるようにスライドする
            while k0 + k1 > W:
                k1, v1 = next(it)
            yield v0 + v1

    # ２つの集合に分けて、それぞれ全列挙
    T0 = make(S[:b//2])
    T1 = make(S[b//2:])

    # 2つの全列挙を元に解を求める
    res = max(solve(T0, T1))
    return res

for _ in range(q):
  x, l = map(int, input().split())
  ans = f(x, l)
  print(ans)