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: Wrong Answer)

use std::cmp::{max, min};
use std::io::{stdin, Read};

#[allow(unused_macros)]
macro_rules! parse {
    ($it: ident ) => {};
    ($it: ident, ) => {};
    ($it: ident, $var:ident : $t:tt $($r:tt)*) => {
        let $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
    ($it: ident, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
}

#[allow(unused_macros)]
macro_rules! parse_val {
    ($it: ident, [$t:tt; $len:expr]) => {
        (0..$len).map(|_| parse_val!($it, $t)).collect::<Vec<_>>();
    };
($it: ident, ($($t: tt),*)) => {
        ($(parse_val!($it, $t)),*)
    };
    ($it: ident, u1) => {
        $it.next().unwrap().parse::<usize>().unwrap() -1
    };
    ($it: ident, $t: ty) => {
        $it.next().unwrap().parse::<$t>().unwrap()
    };
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    parse!(
        it,
        n: usize,
        vw: [(usize, usize); n],
        q: usize,
        vl: [(u1, usize); q]
    );
    let nu = min(2usize.pow(9), n);
    let wm = vl.iter().map(|(_, l)| *l).max().unwrap();
    let mut dp = vec![vec![0; wm + 1]; nu];

    for i in 0..nu {
        let (v, w) = vw[i];
        for j in 0..(wm + 1) {
            if i == 0 {
                if j >= w {
                    dp[i][j] = v;
                }
            } else {
                let pi = (i - 1) / 2;
                dp[i][j] = max(if j >= w { dp[pi][j - w] + v } else { 0 }, dp[pi][j]);
            }
        }
    }

    let mut vls = vec![(0usize, 0usize)];
    let mut buf = vec![];
    for &(i, l) in vl.iter() {
        if i < nu {
            println!("{}", dp[i][l]);
        } else {
            let mut u = i;
            while u >= nu {
                let (vu, lu) = vw[u];
                for &(v, l) in vls.iter() {
                    buf.push((v + vu, l + lu));
                }
                u = (u - 1) / 2;
            }
            vls.append(&mut buf);

            vls.clear();
            vls.push((0, 0));
            buf.clear();

            let mut ret = dp[u][l];
            for &(v, w) in vls.iter() {
                if l >= w {
                    ret = max(ret, dp[u][l - w] + v)
                }
            }
            println!("{}", ret);
        }
    }
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_input() {
        let s = "
        ";
        solve(s);
    }
}