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: Accepted)

use proconio::*;

const MAX: usize = 1 << 18;

struct Solver {
    item: Vec<(u32, usize)>,
    query: Vec<Vec<(usize, usize)>>,
    ans: Vec<u32>,
}

impl Solver {
    fn dfs1(&mut self, v: usize, mut dp: Vec<u32>) {
        let (a, b) = self.item[v];
        for i in (b..dp.len()).rev() {
            dp[i] = std::cmp::max(dp[i], dp[i - b] + a)
        }
        for &(w, k) in self.query[v].iter() {
            self.ans[k] = dp[w];
        }
        if 2 * v >= (1 << 9) {
            self.dfs2(2 * v, &dp, vec![(0, 0)]);
            self.dfs2(2 * v + 1, &dp, vec![(0, 0)]);
        } else {
            self.dfs1(2 * v, dp.clone());
            self.dfs1(2 * v + 1, dp);
        }
    }
    fn dfs2(&mut self, v: usize, dp: &[u32], mut p: Vec<(u32, usize)>) {
        let (a, b) = self.item[v];
        for i in 0..p.len() {
            let (x, y) = p[i];
            p.push((x + a, y + b));
        }
        for &(w, k) in self.query[v].iter() {
            let mut ans = 0;
            for &(a, b) in p.iter().filter(|p| p.1 <= w) {
                ans = std::cmp::max(ans, a + dp[w - b]);
            }
            self.ans[k] = ans;
        }
        if v >= MAX / 2 {
            return;
        } else {
            self.dfs2(2 * v, dp, p.clone());
            self.dfs2(2 * v + 1, dp, p);
        }
    }
}

#[fastout]
fn run() {
    input! {
        n: usize,
        mut item: [(u32, usize); n],
        q: usize,
        ask: [(usize, usize); q],
    }
    while item.len() < MAX {
        item.push((0, 0));
    }
    item.insert(0, (0, 0));
    let mut query = vec![vec![]; MAX + 1];
    for (i, &(v, w)) in ask.iter().enumerate() {
        query[v].push((w, i));
    }
    let mut solver = Solver {
        item: item,
        query: query,
        ans: vec![0; q],
    };
    solver.dfs1(1, vec![0; 100000 + 1]);
    for a in solver.ans {
        println!("{}", a);
    }
}

fn main() {
    run();
}