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)

use std::cmp::*;
use std::io::*;

fn prod(vw: &[(usize, usize)]) -> Vec<(usize, usize)> {
    let mut x = vec![(0, 0)];
    for &(v, w) in vw {
        let mut y = vec![];
        for &(vv, ww) in &x {
            y.push((v + vv, w + ww));
        }
        x.extend(y);
    }
    x
}

trait Bisect: Copy {
    fn halve(self, other: Self) -> Self;
    fn section_end(self, other: Self) -> bool;
}
impl Bisect for i64 {
    fn halve(self, other: Self) -> Self {
        (self + other) / 2
    }
    fn section_end(self, other: Self) -> bool {
        (self - other).abs() <= 1
    }
}
fn binary_search<T: Bisect>(f: &Fn(T) -> bool, ok: T, err: T) -> T {
    let mut ok = ok;
    let mut err = err;
    while !ok.section_end(err) {
        let m = ok.halve(err);
        if f(m) {
            ok = m;
        } else {
            err = m;
        }
    }
    ok
}
fn solve(n: usize, mw: usize, vw: Vec<(usize, usize)>) -> usize {
    let vw1 = prod(&vw[0..n / 2]);
    let mut vw2 = prod(&vw[n / 2..n]);
    vw2.sort_by_key(|&(_, w)| w);
    for i in 0..(vw2.len() - 1) {
        vw2[i + 1].0 = max(vw2[i + 1].0, vw2[i].0);
    }
    let mut ans = 0;
    for &(v, w) in &vw1 {
        let i = binary_search(&|x| mw >= vw2[x as usize].1 + w, -1i64, vw2.len() as i64);
        if i >= 0 {
            ans = max(ans, v + vw2[i as usize].0);
        }
    }
    ans
}

fn main() {
    let mut s: String = String::new();
    stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let mut vw = vec![(0, 0); n + 1];
    for i in 1..=n {
        let a: usize = itr.next().unwrap().parse().unwrap();
        let b: usize = itr.next().unwrap().parse().unwrap();
        vw[i] = (a, b);
    }
    let q: usize = itr.next().unwrap().parse().unwrap();

    let mut out = Vec::new();
    for _ in 0..q {
        let mut v: usize = itr.next().unwrap().parse().unwrap();
        let l: usize = itr.next().unwrap().parse().unwrap();
        let mut y = Vec::new();
        while v > 0 {
            y.push(vw[v]);
            v = v / 2;
        }
        let ans = solve(y.len(), l, y);

        writeln!(out, "{}", ans).ok();
    }
    stdout().write_all(&out).unwrap();
}