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: Time Limit Exceeded)

#![allow(unused_imports, unused_macros, dead_code)]

macro_rules! min {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        it.next().map(|z| it.fold(z, |x, y| min!(x, y)))
    }};
    ($x:expr) => ($x);
    ($x:expr, $($ys:expr),*) => {{
        let t = min!($($ys),*);
        if $x < t { $x } else { t }
    }}
}
macro_rules! max {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        it.next().map(|z| it.fold(z, |x, y| max!(x, y)))
    }};
    ($x:expr) => ($x);
    ($x:expr, $($ys:expr),*) => {{
        let t = max!($($ys),*);
        if $x > t { $x } else { t }
    }}
}
macro_rules! ewriteln {
    ($($args:expr),*) => { let _ = writeln!(&mut std::io::stderr(), $($args),*); };
}
macro_rules! trace {
    ($x:expr) => { ewriteln!(">>> {} = {:?}", stringify!($x), $x) };
    ($($xs:expr),*) => { trace!(($($xs),*)) }
}
macro_rules! put {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        if let Some(x) = it.next() { print!("{}", x); }
        for x in it { print!(" {}", x); }
        println!("");
    }};
    ($x:expr) => { println!("{}", $x) };
    ($x:expr, $($xs:expr),*) => { print!("{} ", $x); put!($($xs),*) }
}

const M: i64 = 1_000_000_007;

fn main() {
    let mut sc = Scanner::new();
    let n: usize = sc.cin();
    let items: Vec<(i64, usize)> = (0..n)
        .map(|_| {
            let v: i64 = sc.cin();
            let w: usize = sc.cin();
            (v, w)
        })
        .collect();

    let q: usize = sc.cin();
    let qs: Vec<(usize, usize)> = (0..q)
        .map(|_| {
            let v: usize = sc.cin();
            let limit: usize = sc.cin();
            (v - 1, limit)
        })
        .collect();

    let mut q_map = vec![vec![]; n];
    for &(v, limit) in qs.iter() {
        q_map[v].push(limit);
    }

    let mut q_memo = std::collections::HashMap::new();

    const W: usize = 100_003;

    let mut stack = vec![];
    let memo = vec![0; W];
    stack.push((0, memo));

    while let Some((v, memo)) = stack.pop() {
        if v >= n {
            continue;
        }
        let mut memo2 = memo.clone();
        for w in (0..W).rev() {
            if w + items[v].1 >= W {
                continue;
            }
            memo2[w + items[v].1] = max!(memo2[w + items[v].1], memo2[w] + items[v].0);
        }
        for w in 1..W {
            memo2[w] = max!(memo2[w], memo2[w - 1]);
        }
        for &limit in q_map[v].iter() {
            // trace!((v, limit), memo2[limit]);
            q_memo.insert((v, limit), memo2[limit]);
        }
        stack.push((v * 2 + 1, memo2.clone()));
        stack.push((v * 2 + 2, memo2.clone()));
    }

    for &(v, limit) in qs.iter() {
        put!(q_memo.get(&(v, limit)).unwrap());
    }
}

use std::collections::VecDeque;
use std::io::{self, Write};
use std::str::FromStr;

struct Scanner {
    stdin: io::Stdin,
    buffer: VecDeque<String>,
}
impl Scanner {
    fn new() -> Self {
        Scanner {
            stdin: io::stdin(),
            buffer: VecDeque::new(),
        }
    }
    fn cin<T: FromStr>(&mut self) -> T {
        while self.buffer.is_empty() {
            let mut line = String::new();
            let _ = self.stdin.read_line(&mut line);
            for w in line.split_whitespace() {
                self.buffer.push_back(String::from(w));
            }
        }
        self.buffer.pop_front().unwrap().parse::<T>().ok().unwrap()
    }
    fn chars(&mut self) -> Vec<char> {
        self.cin::<String>().chars().collect()
    }
    fn vec<T: FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.cin()).collect()
    }
}