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)

use std::cmp;
use std::cmp::Reverse;
use std::collections::BTreeMap;

fn main() {
    let (r, w) = (std::io::stdin(), std::io::stdout());
    let mut sc = IO::new(r.lock(), w.lock());

    let n: usize = sc.read();
    let mut values: Vec<u64> = vec![0; n];
    let mut weights: Vec<u64> = vec![0; n];
    for i in 0..n {
        values[i] = sc.read();
        weights[i] = sc.read();
    }

    let q: usize = sc.read();
    for _ in 0..q {
        let mut v = sc.read::<usize>() - 1;
        let l: u64 = sc.read();
        let mut tv = vec![];
        let mut tw = vec![];
        loop {
            tv.push(values[v]);
            tw.push(weights[v]);
            if v == 0 {
                break;
            }
            v -= 1;
            v /= 2;
        }

        let n = tv.len();
        let head = n / 2;
        let tail = n - head;
        let mut prefix = vec![];
        for mask in 0..(1 << head) {
            let mut v = 0;
            let mut w = 0;
            for i in 0..head {
                if (1 << i) & mask == 0 {
                    continue;
                }
                v += tv[i];
                w += tw[i];
            }
            if w > l {
                continue;
            }
            prefix.push((w, v));
        }
        prefix.sort_by_key(|&(w, v)| (w, Reverse(v)));

        let mut prefix2: Vec<(u64, u64)> = vec![(0, 0)];
        for (w, v) in prefix.into_iter() {
            if w > l {
                continue;
            }
            let prev_v = prefix2[prefix2.len() - 1].1;
            if v <= prev_v {
                continue;
            }
            prefix2.push((w, v));
        }

        let mut suffix = vec![];
        for mask in 0..(1 << tail) {
            let mut v = 0;
            let mut w = 0;
            for i in 0..tail {
                if (1 << i) & mask == 0 {
                    continue;
                }
                v += tv[i + head];
                w += tw[i + head];
            }
            if w > l {
                continue;
            }
            suffix.push((w, v));
        }
        suffix.sort_by_key(|&(w, v)| (w, Reverse(v)));
        let mut suffix2: Vec<(u64, u64)> = vec![(0, 0)];
        for (w, v) in suffix.into_iter() {
            if w > l {
                continue;
            }
            let prev_v = suffix2[suffix2.len() - 1].1;
            if v <= prev_v {
                continue;
            }
            suffix2.push((w, v));
        }

        let mut ans = 0;
        let mut suffix_tail = suffix2.len() - 1;
        for (pw, pv) in prefix2.into_iter() {
            while suffix_tail > 0 && suffix2[suffix_tail].0 + pw > l {
                suffix_tail -= 1;
            }
            if suffix2[suffix_tail].0 + pw > l {
                break;
            }
            ans = cmp::max(ans, pv + suffix2[suffix_tail].1);
        }

        sc.write(ans);
        sc.write("\n");
    }
}

pub struct IO<R, W: std::io::Write>(R, std::io::BufWriter<W>);

impl<R: std::io::Read, W: std::io::Write> IO<R, W> {
    pub fn new(r: R, w: W) -> Self {
        Self(r, std::io::BufWriter::new(w))
    }
    pub fn write<S: ToString>(&mut self, s: S) {
        use std::io::Write;
        self.1.write(s.to_string().as_bytes()).unwrap();
    }
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .0
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r' || b == b'\t')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r' && b != b'\t')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}