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 "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "unordered_map"
#include "unordered_set"
#include "iomanip"
#include "cmath"
#include "random"
#include "bitset"
#include "cstdio"
#include "numeric"
#include "cassert"
#include "ctime"

using namespace std;

constexpr long long int MOD = 1000000007;
//constexpr int MOD = 1000000007;
//constexpr int MOD = 998244353;
//constexpr long long int MOD = 998244353;
constexpr double EPS = 1e-9;

//int N, M, K, H, W, L, R;
long long int N, M, K, H, W, L, R;

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

	cin >> N;
	vector<int>v(N + 1);
	vector<int>w(N + 1);
	for (int i = 1; i <= N; i++) {
		cin >> v[i] >> w[i];
	}
	cin >> K;
	while (K--) {
		L = 200000, R = 100000;
		cin >> L >> R;
		vector<int>V;
		vector<int>W;
		int box = L;
		while (box) {
			V.emplace_back(v[box]);
			W.emplace_back(w[box]);
			box >>= 1;
		}
		int mid = V.size() / 2;
		vector<pair<int, int>>a(1 << mid);
		vector<pair<int, int>>b(1 << (V.size() - mid));
		for (int i = 1; i < 1 << mid; i++) {
			for (int j = 0; j < mid; j++) {
				if ((i >> j) & 1) {
					a[i] = a[i ^ (1 << j)];
					a[i].second += V[j];
					a[i].first += W[j];
					break;
				}
			}
			//	cout << "a " << i << " " << wsum << " " << vsum << endl;
		}
		for (int i = 1; i < 1 << (V.size() - mid); i++) {
			for (int j = 0; j < V.size() - mid; j++) {
				if ((i >> j) & 1) {
					b[i] = b[i ^ (1 << j)];
					b[i].second += V[j + mid];
					b[i].first += W[j + mid];
					break;
				}
			}
			//	cout << "b " << i << " " << wsum << " " << vsum << endl;
		}
		sort(a.begin(), a.end());
		sort(b.begin(), b.end());
		vector<pair<int, int>>c;
		int mx = -1;
		for (auto i : b) {
			if (i.second > mx) {
				c.emplace_back(i);
				mx = i.second;
			}
		}
		reverse(c.begin(), c.end());
		int cnt = 0;
		int ans = 0;
		for (auto i : a) {
			//	cout << i.first << " " << i.second << endl;
			while (cnt + 1 < c.size() && i.first + c[cnt].first > R)cnt++;
			if (i.first +c[cnt].first > R)break;
			ans = max(ans, i.second + c[cnt].second);
		}
		cout << ans << endl;
	}
	return 0;
}