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 Python (3.8.2) (judge verdict: Time Limit Exceeded)

n = int(input())
items = [tuple(map(int, input().split())) for _ in range(n)]

q = int(input())
v_to_qi = [[] for _ in range(n)]
v_to_l = [[] for _ in range(n)]
for i in range(q):
    v, l = map(int, input().split())
    v -= 1
    v_to_qi[v].append(i)
    v_to_l[v].append(l)

MAX = 10**5
# dp[i+1][j] = (i番目まで見て重さの合計がj以下のの時の価値の最大値)
dp = [[0] * (MAX + 1) for _ in range(20)]
ans = [0] * q


def dfs(v, depth):
    for j in range(MAX + 1):
        if j < items[v][1]:
            dp[depth + 1][j] = dp[depth][j]
        else:
            dp[depth + 1][j] = max(dp[depth][j],
                                   dp[depth][j - items[v][1]] + items[v][0])

    if len(v_to_qi[v]) > 0:
        for qi, l in zip(v_to_qi[v], v_to_l[v]):
            ans[qi] = dp[depth + 1][l]

    if v * 2 + 1 < n:
        dfs(v * 2 + 1, depth + 1)
    if v * 2 + 2 < n:
        dfs(v * 2 + 2, depth + 1)


dfs(0, 0)
print(*ans, sep='\n')