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)

def dfs(dpt,mxL,nodes,current,N):
    oya = current//2

    if current > N:
        return
    for omosa,value in dpt[oya].items():
        if omosa > mxL[current]:
            continue
        if omosa in dpt[current]:
            dpt[current][omosa] = max(dpt[current][omosa],value)
        else:
            dpt[current][omosa] = value
        new_omosa = omosa+nodes[current][0]
        if new_omosa > mxL[current]:
            continue
        if new_omosa in dpt[current]:
            dpt[current][new_omosa] = max(dpt[current][new_omosa],value+nodes[current][1])
        else:
            dpt[current][new_omosa] = value+nodes[current][1]

    dfs(dpt,mxL,nodes,current*2,N)
    dfs(dpt,mxL,nodes,current*2+1,N)






def resolve():
    N = int(input())
    nodes =[0] 
    for i in range(N):
        V,W = map(int,input().split())
        nodes.append((W,V))

    q = int(input())
    querys = []
    for i in range(q):
        v,L = map(int,input().split())
        querys.append((v,L))
    mxL = [0]*(N+1)

    for v,L in querys:
         while True:
             mxL[v] = max(mxL[v],L)
             if v == 0:
                 break
             v //=2
    dpt = [dict() for i in range(N+1)]
    dpt[0][0]=0
    dfs(dpt,mxL,nodes,1,N)

    for node, cost in querys:
        mx = 0
        for omosa, value in dpt[node].items():
            if omosa <= cost:
                mx = max(mx,value)
        print(mx)
resolve()