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::input;
use std::cmp::{max, min};

const MAX_L: usize = 100_000;
const BASE_SIZE: usize = 1 << 10;

fn main() {
    input! {
        n: usize,
        vw: [(u32, u32); n],
        q: usize,
        ul: [(usize, usize); q],
    }
    let dp = make_dp(&vw);
    for (u, l) in ul {
        println!("{}", calc(&dp, &vw, u - 1, l));
    }
}

fn make_dp(vw: &Vec<(u32, u32)>) -> Vec<Vec<u32>> {
    let mut dp = Vec::<Vec<u32>>::new();

    dp.push({
        let w = vw[0].1 as usize;
        (0..=MAX_L)
        .map(|i| if i < w { 0 } else { vw[0].0 })
        .collect()
    });

    let n = min(vw.len(), BASE_SIZE - 1);
    for c in 1..n {
        let p = (c - 1) / 2;
        dp.push({
            let w = vw[c].1 as usize;
            let v = vw[c].0;
            (0..=MAX_L)
            .map(|i| {
                max(dp[p][i], if i < w { 0 } else { dp[p][i - w] + v })
            })
            .collect()
        });
    }

    dp
}

fn calc(
    dp: &Vec<Vec<u32>>,
    vw: &Vec<(u32, u32)>,
    u: usize, 
    l: usize,
) -> u32 {
    if u < dp.len() {
        dp[u][l]
    } else {
        let p = (u - 1) / 2;
        let w = vw[u].1 as usize;
        let sum1 = calc(dp, vw, p, l);
        let sum2 = if l < w { 0 } else { calc(dp, vw, p, l - w) + vw[u].0 };
        max(sum1, sum2)
    }
}