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)

def calc_rec(node):
    if node in D.keys():
        return D[node]

    do_list=[node]
    while True:
        node//=2
        if node in D.keys():
            break
        else:
            do_list.append(node)
    for node in reversed(do_list):
        new_dp =calc_dp_step(node)
    return new_dp


def calc_dp_step(node_id):
    cur_dp = D[node_id//2]
    (node_val,node_weight) = nodes[node_id]

    new_dp=[0]*Vmax
    for j in range(Vmax):
        if j < node_weight: 
            new_dp[j] = cur_dp[j]
        else:
            new_dp[j] = max(cur_dp[j],cur_dp[j-node_weight]+node_val)
    D[node_id]=new_dp
    return new_dp

N = int(input())

nodes=[None]
for _ in range(N):
    v,w = map(int,input().split())
    nodes.append((v,w))

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

D=dict()
Vmax=10**5+1

#DP[weight]=val
D[0]=[0]*Vmax
Q_ans=dict()
for v,L in query:
    if v in D.keys():
        new_dp = D[v]
    else:
        new_dp = calc_rec(v)

    print(max(D[v][:L+1]))