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)

macro_rules !read_value {($iter :expr ,($($t :tt ) ,*) ) =>{($(read_value !($iter ,$t ) ) ,*) } ;($iter :expr ,[$t :tt ;$len :expr ] ) =>{(0 ..$len ) .map (|_ |read_value !($iter ,$t ) ) .collect ::<Vec <_ >>() } ;($iter :expr ,{$t :tt =>$f :expr } ) =>{$f (read_value !($iter ,$t ) ) } ;($iter :expr ,chars ) =>{read_value !($iter ,String ) .chars () .collect ::<Vec <char >>() } ;($iter :expr ,usize1 ) =>{read_value !($iter ,usize ) -1 } ;($iter :expr ,$t :ty ) =>{$iter .next () .unwrap () .parse ::<$t >() .unwrap () } ;}
macro_rules !input_inner {($iter :expr ) =>{} ;($iter :expr ,) =>{} ;($iter :expr ,mut $var :ident :$t :tt $($r :tt ) *) =>{let mut $var =read_value !($iter ,$t ) ;input_inner !{$iter $($r ) *} } ;($iter :expr ,mut $var :ident $($r :tt ) *) =>{input_inner !{$iter ,mut $var :usize $($r ) *} } ;($iter :expr ,$var :ident :$t :tt $($r :tt ) *) =>{let $var =read_value !($iter ,$t ) ;input_inner !{$iter $($r ) *} } ;($iter :expr ,$var :ident $($r :tt ) *) =>{input_inner !{$iter ,$var :usize $($r ) *} } ;}
macro_rules !input {(source =$s :expr ,$($r :tt ) *) =>{let mut iter =$s .split_whitespace () ;input_inner !{iter ,$($r ) *} } ;(iter =$iter :ident ,$($r :tt ) *) =>{let s ={use std ::io ::Read ;let mut s =String ::new () ;std ::io ::stdin () .read_to_string (&mut s ) .unwrap () ;s } ;let mut $iter =s .split_whitespace () ;input_inner !{$iter ,$($r ) *} } ;($($r :tt ) *) =>{let s ={use std ::io ::Read ;let mut s =String ::new () ;std ::io ::stdin () .read_to_string (&mut s ) .unwrap () ;s } ;let mut iter =s .split_whitespace () ;input_inner !{iter ,$($r ) *} } ;}
#[macro_export]
macro_rules !min {($e :expr ) =>{$e } ;($e :expr ,$($es :expr ) ,+) =>{std ::cmp ::min ($e ,min !($($es ) ,+) ) } ;}
#[macro_export]
macro_rules !chmin {($dst :expr ,$($src :expr ) ,+) =>{{let x =std ::cmp ::min ($dst ,min !($($src ) ,+) ) ;$dst =x ;} } ;}
#[macro_export]
macro_rules !max {($e :expr ) =>{$e } ;($e :expr ,$($es :expr ) ,+) =>{std ::cmp ::max ($e ,max !($($es ) ,+) ) } ;}
#[macro_export]
macro_rules !chmax {($dst :expr ,$($src :expr ) ,+) =>{{let x =std ::cmp ::max ($dst ,max !($($src ) ,+) ) ;$dst =x ;} } ;}
fn main() {
    #[allow(unused_imports)]
    use std::io::Write as _;
    let __out = std::io::stdout();
    #[allow(unused_mut, unused_variables)]
    let mut __out = std::io::BufWriter::new(__out.lock());
    #[allow(unused_macros)]
    macro_rules !print {($($arg :tt ) *) =>(::std ::write !(__out ,$($arg ) *) .unwrap () ) }
    #[allow(unused_macros)]
    macro_rules !println {($($arg :tt ) *) =>(::std ::writeln !(__out ,$($arg ) *) .unwrap () ) }
    #[allow(unused_macros)]
    macro_rules! echo {
        ($iter :expr ) => {
            echo!($iter, "\n")
        };
        ($iter :expr ,$sep :expr ) => {
            let mut iter = $iter;
            if let Some(item) = iter.next() {
                print!("{}", item);
            }
            for item in iter {
                print!("{}{}", $sep, item);
            }
            println!();
        };
    }
    input!(n, vw: [(u32, u32); n], q, vl: [(usize, u32); q]);
    let mut arr: Vec<Vec<_>> = vec![vec![]; n / 2 + 1];
    arr[0].push((0u32, 0u32));
    for i in 1..=n / 2 {
        let ln = arr[i / 2].len();
        arr[i].reserve(ln * 2);
        for j in 0..ln {
            let (ws, vs) = arr[i / 2][j];
            arr[i].push((ws, vs));
            arr[i].push((ws + vw[i - 1].1, vs + vw[i - 1].0));
        }
        arr[i].sort();
        for j in 1..arr[i].len() {
            chmax!(arr[i][j].1, arr[i][j - 1].1);
        }
    }
    for (u, l) in vl.into_iter() {
        if u <= n / 2 {
            let mut ans = 0;
            for i in 0..arr[u].len() {
                if arr[u][i].0 <= l {
                    chmax!(ans, arr[u][i].1);
                } else {
                    break;
                }
            }
            println!("{}", ans);
        } else {
            let mut xs = vec![(0u32, 0u32)];
            let mut p = u;
            while p > n / 2 {
                for i in 0..xs.len() {
                    let (ws, vs) = xs[i];
                    xs.push((ws + vw[p - 1].1, vs + vw[p - 1].0));
                }
                p = p / 2;
            }
            xs.sort();
            for j in 1..xs.len() {
                chmax!(xs[j].1, xs[j - 1].1);
            }
            let ys = &arr[p];
            let xn = xs.len();
            let yn = ys.len();
            let mut j = yn - 1;
            let mut ans = 0;
            'o: for i in 0..xn {
                while xs[i].0 + ys[j].0 > l {
                    if j == 0 {
                        break 'o;
                    }
                    j -= 1;
                }
                chmax!(ans, xs[i].1 + ys[j].1);
            }
            println!("{}", ans);
        }
    }
}