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 Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::*;

fn calc(mut a: Vec<(u32, i32)>, item: (u32, i32)) -> Vec<(u32, i32)> {
    let mut b = a.iter().map(|a| (a.0 + item.0, a.1 + item.1)).collect::<Vec<_>>();
    let mut c = Vec::with_capacity(a.len() * 2);
    while !(a.is_empty() || b.is_empty()) {
        if a.last().unwrap().1 >= b.last().unwrap().1 {
            c.push(a.pop().unwrap());
        } else {
            c.push(b.pop().unwrap());
        }
    }
    for p in a.into_iter().chain(b.into_iter()).rev() {
        c.push(p);
    }
    c.reverse();
    c
}

#[fastout]
fn run() {
    input! {
        n: usize,
        mut item: [(u32, i32); n],
        q: usize,
        ask: [(usize, i32); q],
    }
    item.insert(0, (0, 0));
    let mut memo = vec![vec![]; n + 1];
    memo[1] = vec![(0, 0), item[1]];
    let batch = 11;
    for i in 2..=std::cmp::min(n, 1 << batch) {
        memo[i] = calc(memo[i / 2].clone(), item[i]);
    }
    for (start, width) in ask {
        let ans = if start < (1 << batch) {
            memo[start].iter().filter(|p| p.1 <= width).max().unwrap().0
        } else {
            let mut dp = vec![(0, 0)];
            let mut pos = start;
            while pos >= (1 << batch) {
                dp = calc(dp, item[pos]);
                pos /= 2;
            }
            let memo = &memo[pos];
            let mut ans = 0;
            let mut val = 0;
            let mut j = memo.len();
            for &(v, w) in dp.iter() {
                while j > 0 && w + memo[j - 1].1 > width {
                    j -= 1;
                }
                val = std::cmp::max(val, v);
                if j > 0 {
                    ans = std::cmp::max(ans, val + memo[j - 1].0);
                }
            }
            ans
        };
        println!("{}", ans);
    }
}

fn main() {
    run();
}