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 <map>
#include <iostream>
using namespace std;
int main() {
    int n; cin >> n;
    int v[n+1], w[n+1];
    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];

    const int S = 2048;
    map<int, long long> mp[S];
    mp[1][0] = 0;
    mp[1][w[1]] = v[1];
    for (int i = 2; i <= min(n, S-1); i++) {
        mp[i] = mp[i/2];
        for (auto it = mp[i/2].rbegin(); it != mp[i/2].rend(); it++) {
            int j = it->first + w[i];
            if (j <= 1e5) {
                mp[i][j] = max(mp[i][j], it->second + v[i]);
            }
        }
        // for (auto x : mp[i]) cerr << x.first << " " << x.second << endl;
    }

    // int q; cin >> q;
    // while (q--) {
    //     int i, l; cin >> i >> l;
    //     map<int, long long> temp;
    //     temp[0] = 0;
    //     while (i >= S) {
    //         for (auto it = temp.rbegin(); it != temp.rend(); it++) {
    //             int j = it->first + w[i];
    //             if (j <= l)
    //                 temp[j] = max(temp[j], it->second + v[i]);
    //         }
    //         i /= 2;
    //     }
    //     //cerr << temp.size() << endl;
    //     long long ans = 0;
    //     auto it1 = mp[i].begin();
    //     auto it2 = temp.rbegin();
    //     while (it1 != mp[i].end() && it2 != temp.rend()) {
    //         while (it2 != temp.rend() && it1->first + it2->first > l) it2++;
    //         if (it2 != temp.rend()) {
    //             ans = max(ans, it1->second + it2->second);
    //         }
    //         it1++;
    //     }
    //     cout << ans << endl;
    // }
}