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)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn main() {
    let n = read::<usize>();
    let mut items = vec![];
    for _ in 0..n {
        let v = read_vec::<usize>();
        let (v, w) = (v[0] as i64, v[1]);
        items.push((w, v));
    }
    let q = read::<usize>();
    let mut queries = vec![];
    for _ in 0..q {
        let v = read_vec::<usize>();
        let (v, l) = (v[0] - 1, v[1]);
        queries.push((v, l));
    }
    let l_max = queries.iter().map(|&x| x.1).max().unwrap();
    let threshold = 511;
    // let threshold = 6;
    let n_dp_max = min(n, threshold);
    let mut dp = vec![vec![-std::i64::MAX; l_max + 1]; n_dp_max];
    dp[0][0] = 0;
    if items[0].0 <= l_max {
        dp[0][items[0].0] = items[0].1;
    }
    for i in 1..n_dp_max {
        let parent = (i - 1) / 2;
        for prev in (0..l_max + 1).rev() {
            if dp[parent][prev] == -std::i64::MAX {
                continue;
            }
            if prev + items[i].0 > l_max {
                continue;
            }
            dp[i][prev + items[i].0] = max(dp[i][prev + items[i].0], dp[parent][prev] + items[i].1);
        }
        for prev in (0..l_max + 1).rev() {
            if dp[parent][prev] == -std::i64::MAX {
                continue;
            }
            dp[i][prev] = max(dp[i][prev], dp[parent][prev]);
        }
    }

    // accum w
    for i in 0..n_dp_max {
        let mut cur = dp[i][0];
        for li in 1..l_max + 1 {
            dp[i][li] = max(dp[i][li], cur);
            cur = dp[i][li];
        }
    }

    for (v, l) in queries {
        let mut cur = v;
        let mut members = vec![];
        while cur >= threshold {
            members.push(cur);
            cur /= 2;
        }
        let mut ans = 0;
        for state in 0..1 << members.len() {
            let mut current_weight = 0;
            let mut current_value = 0;
            for i in 0..members.len() {
                if (state >> i) & 1 == 1 {
                    current_weight += items[members[i]].0;
                    current_value += items[members[i]].1;
                }
            }

            if current_weight > l {
                continue;
            }
            ans = max(ans, current_value + dp[cur][l - current_weight]);
        }
        println!("{}", ans);
    }
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}