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;  // https://docs.rs/proconio/0.4.1/proconio/
use std::cmp::*;

type V = u32;

fn compute_dp(val: &Vec<V>, wgt: &Vec<V>) -> Vec<Vec<V>> {
    let max_l = 100000;
    let n     = val.len();
    let dp_n  = min(n, (1 << 10) - 1);

    let mut dp = Vec::<Vec<V>>::new();
    let row = (0..=max_l)
                  .map(|i| if i < wgt[0] { 0 } else { val[0] })
                  .collect();
    dp.push(row);

    for u in 1..dp_n {
        let p = (u - 1) / 2;
        let row = (0..=max_l)
                      .map(|i| max(dp[p][i as usize],
                                   if i < wgt[u] { 0 } else {
                                       dp[p][(i - wgt[u]) as usize] + val[u]
                                   }))
                      .collect();
        dp.push(row);
    }
    dp
}

fn solve(u: V, l: V, dp: &Vec<Vec<V>>, val: &Vec<V>, wgt: &Vec<V>) -> V
{
    let (uu, ll) = (u as usize, l as usize);
    if uu < dp.len() {
        dp[uu][ll]
    } else {
        let p  = (u - 1) / 2;
        max(solve(p, l,  &dp, &val, &wgt),
            if l < wgt[uu] { 0 } else {
                solve(p, l - wgt[uu], &dp, &val, &wgt) + val[uu]
            })
    }
}

fn main() {
    input! {
        n:       usize,
        val_wgt: [(V, V); n],
        q:       usize,
        query:   [(V, V); q],
    }
    let (val, wgt) = val_wgt.iter().cloned().unzip();

    let dp = compute_dp(&val, &wgt);

    for ul in query.iter() {
        println!("{}", solve(ul.0 - 1, ul.1, &dp, &val, &wgt));
    }
}