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)

#pragma GCC optimize("O2")
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("inline-functions")
#pragma GCC target("sse4,avx2,abm,fma")

#include <cstdio>
#include <algorithm>

typedef std::vector<std::pair<int, int>> vp;

const int N = 1 << 18;

int n, q;
int v[N], w[N];

void dfs(const vp &a, vp &s, int k, int sw, int sv) {
	if (k == (int)a.size()) {
		s.push_back({sw, sv});
		return;
	}
	vp s1, s2;
	dfs(a, s1, k + 1, sw, sv);
	dfs(a, s2, k + 1, sw + a[k].first, sv + a[k].second);
	s.resize(s1.size() + s2.size());
	std::merge(s1.begin(), s1.end(), s2.begin(), s2.end(), s.begin());
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d%d", v + i, w + i);
	}
	scanf("%d", &q);
	while (q--) {
		int u, m;
		scanf("%d%d", &u, &m);
		vp a;
		for (int i = u; i; i /= 2) {
			a.push_back({w[i], v[i]});
		}
		int B = (int)a.size() / 2;
		vp s1, s2;
		dfs(vp(a.begin(), a.begin() + B), s1, 0, 0, 0);
		dfs(vp(a.begin() + B, a.end()), s2, 0, 0, 0);
		for (int i = 1; i < (int)s1.size(); ++i) {
			s1[i].second = std::max(s1[i].second, s1[i - 1].second);
		}
		int ans = 0;
		for (int i = 0, j = (int)s1.size() - 1; i < (int)s2.size(); ++i) {
			while (j >= 0 && s1[j].first + s2[i].first > m) {
				--j;
			}
			if (j >= 0) {
				ans = std::max(ans, s2[i].second + s1[j].second);
			}
		}
		printf("%d\n", ans);
	}
}