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::io::*;

fn main() {
    let mut s: String = String::new();
    stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let mut vw = vec![(0, 0); n + 1];
    for i in 1..=n {
        let a: usize = itr.next().unwrap().parse().unwrap();
        let b: usize = itr.next().unwrap().parse().unwrap();
        vw[i] = (a, b);
    }
    let q: usize = itr.next().unwrap().parse().unwrap();

    let mut out = Vec::new();
    for _ in 0..q {
        let mut v: usize = itr.next().unwrap().parse().unwrap();
        let l: usize = itr.next().unwrap().parse().unwrap();
        let mut y = Vec::new();
        while v > 0 {
            y.push(vw[v]);
            v = v / 2;
        }

        // ここからナップサック
        let n = y.len();
        let m = n / 2;
        let mut ans = 0;
        let mut dp = vec![0; l + 1];
        for i in 0..m {
            for j in (y[i].1..=l).rev() {
                dp[j] = max(dp[j], dp[j - y[i].1] + y[i].0);
            }
        }

        for i in 0..1 << (n - m) {
            let mut tv = 0;
            let mut tw = 0;
            for j in 0..n - m {
                if i >> j & 1 == 1 {
                    tw += y[j + m].1;
                    tv += y[j + m].0;
                }
            }

            if tw <= l {
                ans = max(ans, dp[l - tw] + tv);
            }
        }
        writeln!(out, "{}", ans).ok();
    }
    stdout().write_all(&out).unwrap();
}