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;
#define rep(i, n) for (int i = 0; i < n; i++)
#define repl(i, l, r) for (int i = (l); i < (r); i++)
#define fi first
#define se second
#define all(x) (x).begin(), (x).end()
#define CST(x) cout << fixed << setprecision(x)
using ll = long long;
const ll MOD = 1000000007;
const int inf = 1e9 + 10;
const ll INF = 4e18 + 10;
int v[300000], w[300000];
void solve() {
	int t, l;
	scanf("%d%d", &t, &l);
	int N = 0;
	vector<pair<int, int>> p;
	while (t) {
		p.push_back({v[t], w[t]});
		t /= 2;
		N++;
	}
	int n = (N + 1) / 2, m = N / 2;
	vector<pair<int, int>> A;
	rep(bit, 1 << n) {
		int nowv = 0, noww = 0;
		rep(i, n) {
			if ((bit >> i) & 1) {
				nowv += p[i].first;
				noww += p[i].second;
			}
		}
		if (noww < l) A.push_back({nowv, noww});
	}
	vector<pair<int, int>> B;
	rep(bit, 1 << m) {
		int nowv = 0, noww = 0;
		rep(i, m) {
			if (bit >> i & 1) {
				nowv += p[n + i].first;
				noww += p[n + i].second;
			}
		}
		if (noww < l) B.push_back({noww, -nowv});
	}
	sort(all(B));
	int S = A.size(), T = B.size();
	vector<int> P(T);
	rep(i, T) {
		P[i] = B[i].first;
		if (i)
			B[i].second = max(-B[i].second, B[i - 1].second);
		else
			B[i].second = -B[i].second;
	}
	int ans = 0;
	rep(i, S) {
		int c = upper_bound(all(P), l - A[i].second) - P.begin() - 1;
		ans = max(ans, A[i].first + B[c].second);
	}
	printf("%d\n", ans);
	return;
}
int main() {
	int n;
	scanf("%d", &n);
	rep(i, n) scanf("%d%d", &v[i + 1], &w[i + 1]);
	int q;
	scanf("%d", &q);
	while (q--) solve();
	return 0;
}