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;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

const MX_W: i32 = 100_000;

use std::collections::BTreeMap;

fn dfs(t: &Vec<(i32, i32)>, dp: &mut Vec<Option<BTreeMap<i32, i32>>>, v: usize) {
    if dp[v].is_some() {
        return;
    }
    let p = (v + 1) / 2 - 1;
    if dp[p].is_none() {
        dfs(t, dp, p);
    }

    let r = dp[p].as_ref().unwrap();
    let mut x = r.clone();
    for (w, y) in r {
        if w + t[v].1 > MX_W {
            break;
        }
        let z = x.entry(w + t[v].1).or_insert(0);
        *z = std::cmp::max(*z, y + t[v].0);
    }
    let mut mx = 0;
    for (_, x) in &mut x {
        mx = std::cmp::max(mx, *x);
        *x = mx;
    }
    dp[v] = Some(x);
}

fn main() {
    input! {
        n: usize,
        vw: [(i32, i32); n],
        q: usize,
        vl: [(Usize1, i32); q],
    }

    let mut dp = vec![None; n];
    let mut v = BTreeMap::new();
    v.insert(0, 0);
    v.insert(vw[0].1, vw[0].0);
    dp[0] = Some(v);

    for (v, l) in vl {
        dfs(&vw, &mut dp, v);
        let r = dp[v].as_ref().unwrap();
        println!("{}", r.range(..=l).rev().next().unwrap().1);
    }
}