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 PHP (7.4.4) (judge verdict: Wrong Answer)

<?php
function m($v, $w, $list, $i, $j) {
	if ($i < 0 || $j < 0 || ($i == 0 && $j < $w[$list[0]])) return [0, 0];
	$pre1 = m($v, $w, $list, $i - 1, $j);
	if ($pre1[1] + $w[$list[$i]] <= $j) return [$pre1[0] + $v[$list[$i]], $pre1[1] + $w[$list[$i]]];
	$pre2 = m($v, $w, $list, $i, $j - $w[$list[$i]]);
	return $pre2[0] + $v[$list[$i]] > $pre1[0] ? [$pre2[0] + $v[$list[$i]], $pre2[1] + $w[$list[$i]]] : $pre1;
}
$n = trim(fgets(STDIN));
$v = [];
$w = [];
for ($i = 0; $i < $n; $i++) {
	list($v[$i], $w[$i]) = explode(' ', trim(fgets(STDIN)));
}
$q = trim(fgets(STDIN));
for ($i = 0; $i < $q; $i++) {
	list($x, $l) = explode(' ', trim(fgets(STDIN)));
	$list = [];
	for (; $x >= 1; $x = floor($x / 2)) {
		$list[] = $x - 1;
	}
	echo  m($v, $w, $list, count($list) - 1, $l)[0] . "\n";
}