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 proconio::input;

fn calc(start: usize, end: usize, v: &Vec<(i64, i64)>) -> Vec<(i64, i64)> {
    let mut res = Vec::new();
    let mut val = Vec::new();
    let mut idx = start;
    while idx > end {
        val.push(v[idx-1]);
        idx /= 2;
    }
    for i in 0..1<<val.len() {
        let mut sv = 0;
        let mut sw = 0;
        for j in 0..val.len() {
            if i&(1<<j) != 0 {
                sv += val[j].0;
                sw += val[j].1;
            }
        }
        res.push((sw, sv));
    }
    res.sort();
    for i in 1..res.len() {
        res[i].1 = std::cmp::max(res[i].1, res[i-1].1);
    }
    res
}

fn main() {
    input! {
        n : usize,
        v : [(i64, i64); n],
        q : usize,
        query : [(usize, i64); q]
    };
    let thr = 512;
    let mut first = Vec::new();
    for i in 0..std::cmp::min(thr, n) {
        first.push(calc(i+1, 0, &v));
    }
    for &t in &query {
        let mut res = 0;
        if t.0 <= thr {
            for &r in &first[t.0-1] {
                if r.0 > t.1 { continue }
                res = std::cmp::max(res, r.1);
            }
        } else {
            let second = calc(t.0, thr, &v);
            let mut par = t.0;
            while par > thr { par /= 2; }
            par -= 1;
            let mut idx0 = 0;
            let mut idx1 = second.len() - 1;
            while idx0 < first[par].len() && idx1 < second.len() {
                if first[par][idx0].0 + second[idx1].0 <= t.1 {
                    res = std::cmp::max(res, first[par][idx0].1 + second[idx1].1);
                    idx0 += 1;
                } else {
                    idx1 = idx1.wrapping_add(!0);
                }
            }
        }
        println!("{}", res);
    }
}