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: Time Limit Exceeded)

#include <iostream>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;
typedef long long ll;
int N;
int Q;
ll W[(1<<18)+1], V[(1<<18)+1];
ll dp[(1<<9)+1][100001];
ll dp_prev[100001];
ll dp_cur[100001];

int D;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(10) << fixed;
    cin >> N;
    for(int i = 1; i <= N; i++) cin >> V[i] >> W[i];
    for(int i = 0; i < 20; i++){
        if(!((1<<i)&N)){
            D = i;
            break;
        }
    }
    cin >> Q;
    int M = D/2;
    for(int i = 1; i < (1<<M); i++){
        for(int i = 0; i <= 100000; i++){
            dp_cur[i] = 0; dp_prev[i] = 0;
        }
        int cur = i;
        while(true){
            for(int j = 0; j+W[cur] <= 100000; j++){
                dp_cur[j+W[cur]] = max(dp_cur[j+W[cur]], dp_prev[j]+V[cur]);
            }
            for(int j = 0; j <= 100000; j++) dp_prev[j] = dp_cur[j];
            cur = cur/2;
            if(cur == 0) break;
        }
        for(int j = 1; j <= 100000; j++) {
            dp[i][j] = max(dp[i][j-1], dp_cur[j]);
        }
    }
    int th = (1<<M);
    for(int i = 0; i < Q; i++){
        int v, L;
        cin >> v >> L;
        if(v < th){
            cout << dp[v][L] << endl;
            continue;
        }
        int cur = v;
        vector<int> vv, vw;
        while(true){
            vv.push_back(V[cur]);
            vw.push_back(W[cur]);
            cur = cur/2;
            if(cur < th) break;
        }
        int sz = vv.size();
        // cout << i << " cur:" << cur << ' ' << sz << endl;
        int ans = 0;
        for(int j = 0; j < (1<<sz); j++){
            int sumv = 0, sumw = 0;
            for(int k = 0; k < sz; k++){
                if(j&(1<<k)){
                    sumv += vv[k];
                    sumw += vw[k];
                    // cout  << vv[k] << '-' << vw[k] << ' ';
                }
            }
            if(sumw > L) continue;
            int tmp = sumv+dp[cur][L-sumw];
            ans = max(ans, tmp);
            // cout << sumv << ' ' << sumw << ' ' << tmp << ' ' << ans<< ' ' << dp[cur][L-sumw] << ' ' << L-sumw << endl;
        }
        cout << ans << endl;
    }
}