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: Accepted)

#include <bits/stdc++.h>

#define ll long long
#define ld long double
#define eb emplace_back
#define fi first
#define se second

using namespace std;

const int nm = (1 << 18) + 2;
const int mm = 100002;
const int nm2 = (1 << 9) + 2;

int n, w[nm], v[nm];
int f[nm2][mm];
int m;

int solve(int u, int l) {
	if (u < nm2) {
		return f[u][l];
	}
	int res = solve(u >> 1, l);
	if (w[u] <= l) {
		res = max(res, solve(u >> 1, l - w[u]) + v[u]);
	}
	return res;
}

int main() {
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
#endif
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d%d", &v[i], &w[i]);
	}
	for (int i = 1; i < nm2 && i <= n; ++i) {
		for (int j = 1; j < mm; ++j) {
			f[i][j] = f[i >> 1][j];
			if (w[i] <= j) {
				f[i][j] = max(f[i][j], f[i >> 1][j - w[i]] + v[i]);
			}
		}
	}
	scanf("%d", &m);
	while (m--) {
		int u, l;
		scanf("%d%d", &u, &l);
		printf("%d\n", solve(u, l));
	}
#ifdef LOCAL
	cout << setprecision(6) << fixed << (double) clock() / CLOCKS_PER_SEC << "\n";
#endif
}