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 proconio::{input, marker::*};

#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};

#[allow(unused_imports)]
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        #[cfg(debug_assertions)]
        writeln!(&mut std::io::stderr(), concat!("[DEBUG] ", $(stringify!($a), "={:?} "),*), $($a),*).unwrap();
    }
}

macro_rules ! max { ( $ x : expr ) => ( $ x ) ; ( $ x : expr , $ ( $ z : expr ) ,+ ) => { { std :: cmp :: max ( $ x , max ! ( $ ( $ z ) ,* ) ) } } }
macro_rules ! min { ( $ x : expr ) => ( $ x ) ; ( $ x : expr , $ ( $ z : expr ) ,+ ) => { { std :: cmp :: min ( $ x , min ! ( $ ( $ z ) ,* ) ) } } }

const HALF: usize = 4096;
const MAX: usize = 100_001;

fn dfs(n: usize, i: usize, dp: &mut Vec<Vec<usize>>, vw: &Vec<(usize, usize)>) {
    let (v, w) = vw[i];
    if i == 0 {
        for j in 0..MAX {
            dp[i][j] = if j < w { 0 } else { v };
        }
    } else {
        let p = (i - 1) / 2;
        for j in 0..MAX {
            dp[i][j] = if j < w {
                dp[p][j]
            } else {
                max!(dp[p][j], dp[p][j - w] + v)
            };
        }
    }

    for j in 0..2 {
        let next = 2 * i + j + 1;
        if next >= n {
            continue;
        }
        dfs(n, next, dp, vw);
    }
}

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

    let half = min!(n, HALF);

    let mut dp = vec![vec![0; MAX]; half];
    dfs(half, 0, &mut dp, &vw);

    for (v, l) in vl {
        if v < half {
            println!("{}", dp[v][l]);
            continue;
        }

        let mut vec = Vec::new();
        let mut pos = v;
        while pos >= half {
            vec.push(pos);
            pos = (pos - 1) / 2;
        }

        let size = vec.len();
        let mut mx = 0;
        for i in 0..1 << size {
            let mut val = 0;
            let mut wei = 0;
            for j in 0..size {
                if (i >> j) & 1 == 0 {
                    continue;
                }
                let (v, w) = vw[vec[j]];
                val += v;
                wei += w;
            }
            if wei > l {
                continue;
            }
            mx = max!(mx, val + dp[pos][l - wei]);
        }
        println!("{}", mx);
    }
}