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)]
#![allow(non_snake_case, unused)]

use std::cmp::*;
use std::collections::*;
use std::ops::*;

// https://atcoder.jp/contests/hokudai-hitachi2019-1/submissions/10518254

macro_rules! eprint {
	($($t:tt)*) => {{
		use ::std::io::Write;
		let _ = write!(::std::io::stderr(), $($t)*);
	}};
}
macro_rules! eprintln {
	() => { eprintln!(""); };
	($($t:tt)*) => {{
		use ::std::io::Write;
		let _ = writeln!(::std::io::stderr(), $($t)*);
	}};
}
macro_rules! dbg {
	($v:expr) => {{
		let val = $v;
		eprintln!("[{}:{}] {} = {:?}", file!(), line!(), stringify!($v), val);
		val
	}}
}

macro_rules! mat {
	($($e:expr),*) => { Vec::from(vec![$($e),*]) };
	($($e:expr,)*) => { Vec::from(vec![$($e),*]) };
	($e:expr; $d:expr) => { Vec::from(vec![$e; $d]) };
	($e:expr; $d:expr $(; $ds:expr)+) => { Vec::from(vec![mat![$e $(; $ds)*]; $d]) };
}

macro_rules! ok {
	($a:ident$([$i:expr])*.$f:ident()$(@$t:ident)*) => {
		$a$([$i])*.$f($($t),*)
	};
	($a:ident$([$i:expr])*.$f:ident($e:expr$(,$es:expr)*)$(@$t:ident)*) => { {
		let t = $e;
		ok!($a$([$i])*.$f($($es),*)$(@$t)*@t)
	} };
}

pub fn readln() -> String {
	let mut line = String::new();
	::std::io::stdin().read_line(&mut line).unwrap_or_else(|e| panic!("{}", e));
	line
}

macro_rules! read {
	($($t:tt),*; $n:expr) => {{
		let stdin = ::std::io::stdin();
		let ret = ::std::io::BufRead::lines(stdin.lock()).take($n).map(|line| {
			let line = line.unwrap();
			let mut it = line.split_whitespace();
			_read!(it; $($t),*)
		}).collect::<Vec<_>>();
		ret
	}};
	($($t:tt),*) => {{
		let line = readln();
		let mut it = line.split_whitespace();
		_read!(it; $($t),*)
	}};
}

macro_rules! _read {
	($it:ident; [char]) => {
		_read!($it; String).chars().collect::<Vec<_>>()
	};
	($it:ident; [u8]) => {
		Vec::from(_read!($it; String).into_bytes())
	};
	($it:ident; usize1) => {
		$it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1
	};
	($it:ident; [usize1]) => {
		$it.map(|s| s.parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1).collect::<Vec<_>>()
	};
	($it:ident; [$t:ty]) => {
		$it.map(|s| s.parse::<$t>().unwrap_or_else(|e| panic!("{}", e))).collect::<Vec<_>>()
	};
	($it:ident; $t:ty) => {
		$it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<$t>().unwrap_or_else(|e| panic!("{}", e))
	};
	($it:ident; $($t:tt),+) => {
		($(_read!($it; $t)),*)
	};
}


pub fn main() {
	let _ = ::std::thread::Builder::new().name("run".to_string()).stack_size(32 * 1024 * 1024).spawn(run).unwrap().join();
}

const MOD: u32 = 1_000_000_007;
const INF: i64 = std::i64::MAX/3;

fn solve() {
	let n = read!(usize);
	let vw = read!(usize,usize;n);
	let q = read!(usize);
	let vl = read!(usize,usize;q);
	let half = min(1<<9,n);
	let mut dp = vec![vec![0;100001];half];
	for j in vw[0].1..100001 {
		dp[0][j] = vw[0].0;
	}
	for i in 1..half {
		let p = (i+1)/2-1;
		for j in 0..100001 {
			let nw = j+vw[i].1;
			if nw<=100000{
				dp[i][nw] = max(dp[i][nw], dp[p][j]+vw[i].0);
			}
			dp[i][j] = max(dp[i][j],dp[p][j]);
		}
	}
	for (v,l) in vl {
		let mut ans = 0;
		let mut id = v-1;
		let mut list = vec![];
		let mut p = 0;
		loop {
			if id<half {
				p = id;
				break;
			}
			list.push(id);
			id = (id+1)/2-1;
		}
		for j in 0..1<<(list.len()) {
			let mut vsum = 0;
			let mut wsum = 0;
			for k in 0..list.len() {
				if (j>>k)%2==1 {
					vsum += vw[list[k]].0;
					wsum += vw[list[k]].1;
				}
			}
			if wsum > l {
				continue;
			}
			let res = l-wsum;
			ans = max(ans, dp[p][res]+vsum);
		}
		println!("{}",ans);
	}
}

fn run() {
    solve();
}