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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <complex>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <stack> 
#include <bitset>
#include <queue>
#include <assert.h>
using namespace std;

#define x first
#define y second
const int N = 1 << 19;
int a[N], b[N];
int limits = 6;
const int C = 12;
vector<pair<int, int> > v[(1 << C) + 10];
int c[C + 10];
int aa[N], bb[N];

void init(int n) {
    int cnt = 0;
	int m = n;
    while (n >= 1) {
	    c[cnt++] = n;
	    n >>= 1;
	}

	v[m].push_back(make_pair(0, 0));
    for (int i = 1; i < (1 << cnt); i++) {
		int x = 0, y = 0;
		for (int j = 0; j < cnt; j++)
			if (i & (1 << j)) {
				x += a[c[j]];
				y += b[c[j]];
			}

		v[m].push_back(make_pair(y, x));
	}	
	sort(v[m].begin(), v[m].end());
	for (int i = 1; i <= v[m].size(); i++)
		v[m][i].y = max(v[m][i - 1].y, v[m][i].y);
}

void solve() {
	int n, m;
	scanf("%d%d", &n, &m);
	int cnt = 0;
	while (n > (1 << C)) {
		c[cnt++] = n;
		n >>= 1;
	}

	int ans = 0;
	for (int i = 0; i < (1 << cnt); i++) {
		int x = 0, y = 0;
		for (int j = 0; j < cnt; j++)
			if (i & (1 << j)) {
				x += a[c[j]];
				y += b[c[j]];
			}
		int lt = 0, rt = v[n].size() - 1, mid;
		while (lt <= rt) {
			mid = (lt + rt) >> 1;
			if (y + v[n][mid].x <= m) {
				ans = max(ans, x + v[n][mid].y);
		        lt = mid + 1;
			} else rt = mid - 1;
		}
	}
	printf("%d\n", ans);
}

int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d%d", &a[i], &b[i]);
	for (int i = 1; i <= (1 << C); i++)
		init(i);

	int m;
	scanf("%d", &m);
	while (m--) solve();
	return 0;
}