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: Runtime Error)

#![allow(non_snake_case,unused_imports,dead_code)]
use std::io::Read;
use std::cmp::max;
use std::cmp::min;
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::BinaryHeap;
use std::cmp::Ordering;

fn comp_f64(a: &f64, b: &f64) -> Ordering {
    if a < b {
        return Ordering::Less;
    } else if a > b {
        return Ordering::Greater;
    }
    Ordering::Equal
}

fn main() {

    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut buf_it = buf.split_whitespace();


    let N = buf_it.next().unwrap().parse::<usize>().unwrap();
    
    let VW = (0..N).map(|_|
        (buf_it.next().unwrap().parse::<usize>().unwrap(),
        buf_it.next().unwrap().parse::<usize>().unwrap())
    ).collect::<Vec<_>>();

    
    let Q = buf_it.next().unwrap().parse::<usize>().unwrap();
    
    let vL = (0..Q).map(|_|
        (buf_it.next().unwrap().parse::<usize>().unwrap(),
        buf_it.next().unwrap().parse::<usize>().unwrap())
    ).collect::<Vec<_>>();

    let mut L_max = (0..N+1).map(|_| 0).collect::<Vec<_>>();
    
    for &(v,l) in &vL{
        L_max[v] = max(L_max[v],l);
    }
    for i in 0..N{
        let i = N-i;
        let p = i/2;
        L_max[p] = max(L_max[p],L_max[i]);
    }
    let mut W_sum = (0..N+1).map(|_| 0).collect::<Vec<_>>();
    for i in 0..N{
        let w = VW[i].1;
        let i = i + 1;
        let p = i/2;
        W_sum[i] = min(W_sum[p] + w,L_max[i]);
    }

    let mut dp = (0..N+1).map(|i| 
                    (0..(W_sum[i]+1)).map(|_| 0).collect::<Vec<_>>()
                ).collect::<Vec<_>>();

    for i in 1..N+1{
        let p = i/2;
        for j in 0..min(W_sum[i],W_sum[p])+1{
            dp[i][j] = dp[p][j];
        }
        let wsp = W_sum[p];
        for j in W_sum[p]+1..W_sum[i]+1{
            dp[i][j] = dp[p][wsp];
        }
        let (v,w) = VW[i-1];
        for j in 0..W_sum[i]+1-w{
            dp[i][j+w] = max(dp[i][j+w],dp[p][j] + v);
        }
    }

    for &(v,l) in &vL{
        println!("{}",dp[v][l]);
    }

}