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)]
use proconio::{fastout, input, marker::*};

use std::collections::BTreeMap;

#[fastout]
fn main() {
   input! {
      n: usize,
      vw: [(usize, usize); n],
      q: usize,
      vl: [(Usize1, usize); q],
   }

   for (v, l) in vl {
      let mut v = v;
      let mut node = vec![0];
      while v != 0 {
         node.push(v);
         v = (v - 1) / 2;
      }
      let mut dp: BTreeMap<usize, usize> = BTreeMap::new();
      dp.insert(0, 0);
      for j in node {
         let mut dp2: BTreeMap<usize, usize> = dp.clone();
         for (&w, &v) in dp.iter() {
            if w + vw[j].1 <= l {
               dp2.entry(w + vw[j].1)
                  .and_modify(|x| *x = std::cmp::max(*x, v + vw[j].0))
                  .or_insert(v + vw[j].0);
            }
         }
         dp = dp2;
      }
      let mut ans = 0;
      for (_, v) in dp {
         ans = std::cmp::max(ans, v);
      }
      println!("{}", ans);
   }
}