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**8)
input = sys.stdin.readline

N = int(input())
VW = [list(map(int,input().split())) for _ in range (N)]
Q = int(input())
vL = [list(map(int,input().split())) for _ in range (Q)]

# 深さ9まではDPテーブルで計算　(L_max = 10**5+1)
# 親のDPテーブルから遷移させる
L_max = 10**5+1
N_max = min(1<<9, N)
dp = [[0]*L_max for _ in range(N_max+1)]
for i in range(N_max):
    for j in range(L_max):
        if j>=VW[i][1]:
            dp[i+1][j] = max(dp[(i+1)//2][j], dp[(i+1)//2][j-VW[i][1]]+VW[i][0])
        else:
            dp[i+1][j] = dp[(i+1)//2][j]

Th  = 1<<9
for v, L in vL:
    if v <= Th:
        print(dp[v][L])
        continue
    # v>Thの場合，vからThより大きい番号の祖先を使って全列挙
    # 全列挙して，DPテーブルと組み合わせて最大値を計算

    ans = 0
    # 末端の要素でリスト生成
    weights = []
    while v>Th:
        weights.append((VW[v-1][0],VW[v-1][1]))
        v >>= 1
    # bit全探索
    for i in range(1<<(len(weights))):
        tw, tv = 0,0
        for j in range(len(weights)):
            if i==0:
                break
            if i&1:
                tw += weights[i][1]
                tv += weights[i][0]
            i >>= 1

        if tw>L:continue
        tv += dp[v][L-tw]
        ans = max(tv, ans)
    print(ans)