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::*;

// ナップザック？
// 死ぬ
//
// アイテムはlog N 程度 <= 18
// 半分全列挙？
// 各クエリ2^9 程度で処理できればOKだが
//
// 2^18 = 2.6*10^5
//
// 実装スッキリしない

#[fastout]
fn run() {
    input! {
        n: usize,
        mut item: [(u32, i32); n],
        q: usize,
        ask: [(usize, i32); q],
    }
    item.insert(0, (0, 0));
    let calc = |v: usize| -> (usize, Vec<(u32, i32)>) {
        let mut dp = vec![(0, 0)];
        let mut pos = v;
        while pos >= 1 && dp.len() < (1 << 9) {
            let (v, w) = item[pos];
            pos /= 2;
            let mut a = vec![];
            for &(p, q) in dp.iter() {
                a.push((p + v, q + w));
            }
            let mut next = vec![];
            while a.len() > 0 && dp.len() > 0 {
                let (_, q) = *a.last().unwrap();
                let (_, s) = *dp.last().unwrap();
                if q >= s {
                    next.push(a.pop().unwrap());
                } else {
                    next.push(dp.pop().unwrap());
                }
            }
            while let Some(p) = dp.pop() {
                next.push(p);
            }
            while let Some(p) = a.pop() {
                next.push(p);
            }
            next.reverse();
            dp = next;
        }
        let mut v = 0;
        for i in 0..dp.len()  {
            v = std::cmp::max(v, dp[i].0);
            dp[i].0 = v;
        }
        (pos, dp)
    };
    for (start, width) in ask {
        let (v, dp) = calc(start);
        let ans = if v >= 1 {
            let p = calc(v).1;
            let mut ans = 0;
            let mut i = 0;
            let mut j = p.len();
            while i < dp.len() && j > 0 {
                if dp[i].1 + p[j - 1].1 <= width {
                    ans = std::cmp::max(ans, dp[i].0 + p[j - 1].0);
                    i += 1;
                } else {
                    j -= 1;
                }
            }
            ans
        } else {
            let mut ans = 0;
            for (v, w) in dp {
                if w <=  width {
                    ans = std::cmp::max(v, ans);
                }
            }
            ans
        };
        println!("{}", ans);
    }
}

fn main() {
    run();
}