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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
using ll = long long;

template <class T>
void chmax(T &a, T b) {
    if (b > a) a = b;
}

constexpr int W = (int)1e+5;

int dp[1 << 8][W + 1];//それ以下の重さで実現できる最大価値

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n;

    vector<int> v(n + 1), w(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> v[i] >> w[i];
    }

    if (n >= 1 << 9) {
        for (int k = 0; k < 1 << 8; k++) {
            int i = 1;
            for (int h = 0; h < 9; h++) {
                for (int j = W; j >= w[i]; j--) {
                    chmax(dp[k][j], dp[k][j - w[i]] + v[i]);
                }
                i *= 2;
                i += (k >> (8 - h) & 1);
            }
        }
    }

    int q;
    cin >> q;

    while (q--) {
        int i0, L;
        cin >> i0 >> L;

        int i = i0;
        int h0 = 0;
        int l = 0;

        if (i < 1 << 9) {
            for (int h = 1;; h++) {
                i /= 2;
                if (i == 0) {
                    h0 = h;
                    l = -1;
                    break;
                }
            }
        } else {
            for (int h = 1;; h++) {
                i /= 2;
                if (i < 1 << 9) {
                    h0 = h;
                    l = i - (1 << 8);
                    break;
                }
            }
        }

        int r = 0;
        if (i < 1 << 9) {
            for (int k = 0; k < 1 << h0; k++) {
                int i = i0, v1 = 0, w1 = 0;
                for (int h = 0; h < h0; h++) {
                    if (k >> h & 1) {
                        v1 += v[i];
                        w1 += w[i];
                    }
                    i /= 2;
                }
                if (w1 > L) continue;
                chmax(r, v1);
            }
        } else {
            for (int k = 0; k < 1 << h0; k++) {
                int i = i0, v1 = 0, w1 = 0;
                for (int h = 0; h < h0; h++) {
                    if (k >> h & 1) {
                        v1 += v[i];
                        w1 += w[i];
                    }
                    i /= 2;
                }
                if (w1 > L) continue;
                chmax(r, dp[l][L - w1] + v1);
            }
        }

        cout << r << '\n';
    }

    return 0;
}