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 <bits/stdc++.h>

using namespace std;

template <class TH> void _dbg(const char *sdbg, TH h){cerr<<sdbg<<"="<<h<<"\n";}
template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {
  while(*sdbg!=',')
    cerr<<*sdbg++;
  cerr<<"="<<h<<","; 
  _dbg(sdbg+1, a...);
}

template<class T> ostream & operator<<(ostream & os, vector<T> V){
  os<<"[";
  for(auto vv: V) os << vv <<",";
  return os << "]";
}
template<class L, class R> ostream & operator <<(ostream & os, pair<L,R> P){
  return os <<"("<<P.first <<","<<P.second <<")";
}

#ifdef DEBUG
#define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#else
#define debug(...) (__VA_ARGS__)
#define cerr if(0)cout
#endif

const int N = (1 << 18) + 10;

int q, n, value[N], weight[N];
vector<pair<int, int>> vec;
vector<pair<int, int>> countLeft[100005], countRight[100005], sbLeft[100005], sbRight[100005];

void backtrack(vector<pair<int, int>> &items, int pos, int totalWeight, int totalValue, int L, int queryId, int type) {
    if (totalWeight > L) {
        return;
    }
    if (pos == items.size()) {
//         vec.push_back({totalWeight, totalValue});
        if (type == 0)
        countLeft[totalWeight].push_back({totalValue, queryId});
        else {
            
        countRight[totalWeight].push_back({totalValue, queryId});
        }
        return;
    }
    backtrack(items, pos + 1, totalWeight, totalValue, L, queryId, type);
    backtrack(items, pos + 1, totalWeight + items[pos].second, totalValue + items[pos].first, L, queryId, type);
}

void genSubsets(vector<pair<int, int>> items, int L, int queryId, int type) {
    vec.clear();
    sort(items.begin(), items.end(), [&](auto a, auto b) { return a.second > b.second; });
    backtrack(items, 0, 0, 0, L, queryId, type);
}


void solve(int v, int L, int queryId) {
    vector<pair<int, int> > items;
    for (int w = v; w >= 1; w /= 2) {
        if (weight[w] <= L) {
            items.push_back({value[w], weight[w]});
        }
    }
    random_shuffle(items.begin(), items.end());
    int half = items.size() / 2;
    vector<pair<int, int>> left, right;
    for (int i = 0; i < half; i++) {
        left.push_back(items[i]);
    }
    for (int i = half; i < items.size(); i++) {
        right.push_back(items[i]);
    }
    genSubsets(left, L, queryId, 0);
    genSubsets(right, L, queryId, 1);

//     sort(subsetsLeft.begin(), subsetsLeft.end());
//     sort(subsetsRight.begin(), subsetsRight.end());

}

int getAnswer(vector<pair<int, int>> subsetsLeft, vector<pair<int, int>> subsetsRight, int L) {
    for (int i = 1; i < subsetsLeft.size(); i++) {
        subsetsLeft[i].second = max(subsetsLeft[i].second, subsetsLeft[i - 1].second);
    }
    for (int i = 1; i < subsetsRight.size(); i++) {
        subsetsRight[i].second = max(subsetsRight[i].second, subsetsRight[i - 1].second);
    }
    
    int ptr = subsetsRight.size() - 1;
    int ret = 0;
    for (int i = 0; i < subsetsLeft.size(); i++) {
        while (ptr > 0 && subsetsLeft[i].first + subsetsRight[ptr].first > L) {
            ptr--;
        }
        if (subsetsLeft[i].first > L) break;
        if (ptr > 0) {
            ret = max(ret, subsetsLeft[i].second + subsetsRight[ptr].second);
        } else {
            ret = max(ret, subsetsLeft[i].second);
        }
    }
    return ret;
}

int memL[100005];
int main() {
    
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d %d", &value[i], &weight[i]);
    }
    
    scanf("%d", &q);
    for (int i = 1; i <= q; i++) {
        int v, L;
        scanf("%d %d", &v, &L);
        memL[i] = L;
        solve(v, L, i);
    }
    
    for (int i = 0; i <= 100000; i++) {
        for (auto p : countLeft[i]) {
            sbLeft[p.second].push_back({i, p.first});
        }
        for (auto p : countRight[i]) {
            sbRight[p.second].push_back({i, p.first});
        }
    }
    
    for (int i = 1; i <= q; i++) {
        printf("%d\n", getAnswer(sbLeft[i], sbRight[i], memL[i]));
    }
    
    return 0;
}