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: Accepted)

def main():
    import sys
    input = sys.stdin.readline

    n = int(input())

    L_MAX = 10**5
    D_TH = 10
    V_MAX = (1 << D_TH) - 1
    dp = [[0] * (L_MAX + 1)]
    # items = []
    items_v = []
    items_w = []

    for i in range(n):
        val, w = map(int, input().split())
        if i < V_MAX:
            i1 = i + 1
            i0 = (i + 1) >> 1
            dp_ = dp[i0][:]

            # for j in range(w):
            #     dp[i1][j] = dp[i0][j]
            for j in range(w, L_MAX + 1)[::-1]:
                if dp_[j] < dp_[j - w] + val:
                    dp_[j] = dp_[j - w] + val
            dp.append(dp_)
        else:
            # items.append((val, w))
            items_v.append(val)
            items_w.append(w)

    # items = tuple(items)

    q = int(input())
    for _ in range(q):
        v, l = map(int, input().split())
        v -= 1

        if v < V_MAX:
            ans = dp[v + 1][l]
        else:
            vv = []
            ww = []
            while v >= V_MAX:
                # vv.append(items[v - V_MAX][0])
                # ww.append(items[v - V_MAX][1])
                vv.append(items_v[v - V_MAX])
                ww.append(items_w[v - V_MAX])
                v = (v - 1) >> 1

            ans = dp[v + 1][l]
            # m = len(vv)
            m2 = 1 << len(vv)
            v_sub = [0] * m2
            w_sub = [0] * m2
            for bit in range(1, m2):
                bit_min = bit & (-bit)
                x = bit_min.bit_length() - 1

                s_w = w_sub[bit - bit_min] + ww[x]
                w_sub[bit] = s_w
                if s_w <= l:
                    s_val = v_sub[bit - bit_min] + vv[x]
                    v_sub[bit] = s_val
                    ans = max(ans, dp[v + 1][l - s_w] + s_val)

        print(ans)


if __name__ == '__main__':
    main()