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)]
use itertools::Itertools;
#[allow(unused_imports)]
use num::Integer;
#[allow(unused_imports)]
use proconio::derive_readable;
use proconio::fastout;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::cmp;

const W: usize = 100_000;
// const W: usize = 10;

#[fastout]
fn main() {
    input! {
        n: usize,
        item: [(i64, usize); n],
        q: usize,
        query: [(Usize1, usize); q],
    }
    let m = cmp::min(n, 512) as usize;
    let mut max_val = vec![vec![0i64; W + 1]; m + 1];
    for (i, &(a, b)) in item.iter().take(m).enumerate() {
        for w in 0..=W {
            max_val[i + 1][w] =
                cmp::max(max_val[(i + 1) / 2][w], max_val[i + 1][w.saturating_sub(1)]);
            if let Some(pw) = w.checked_sub(b) {
                max_val[i + 1][w] = cmp::max(max_val[i + 1][w], max_val[(i + 1) / 2][pw] + a);
            }
        }
    }
    for &(v, l) in &query {
        if v < m {
            println!("{}", max_val[v + 1][l]);
        } else {
            let mut vs = vec![v];
            let mut v = v;
            loop {
                v = (v - 1) / 2;
                if v < m {
                    break;
                }
                vs.push(v);
            }
            let last_v = v;
            let mut ret = 0i64;
            for pat in 0..1 << vs.len() {
                let mut value = 0i64;
                let mut w = 0usize;
                for (i, &v) in vs.iter().enumerate() {
                    if (pat >> i) & 1 != 0 {
                        value += item[v].0;
                        w += item[v].1;
                    }
                }

                if let Some(pw) = l.checked_sub(w) {
                    ret = cmp::max(ret, value + max_val[last_v + 1][pw]);
                }
            }
            println!("{}", ret);
        }
    }
}