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 OCaml (4.10.0) (judge verdict: Time Limit Exceeded)

let rec range a b = if a >= b then Seq.empty else fun () -> Seq.Cons (a, range (a+1) b)

let n = Scanf.scanf "%d" Fun.id
let vw = List.init n (fun _ -> Scanf.scanf " %d %d" (fun v w -> v, w))
let varr = List.to_seq vw |> Seq.map fst |> Array.of_seq
let warr = List.to_seq vw |> Seq.map snd |> Array.of_seq

let rec items arr x =
  if x = 1 then [arr.(0)]
  else arr.(x-1) :: items arr (x / 2)

let enumerate t lim va wa =
  let ve = Array.make (1 lsl t) 0 in
  let we = Array.make (1 lsl t) 0 in
  range 0 (1 lsl t) |> Seq.iter (fun b ->
    let ws = range 0 t |> Seq.fold_left (fun ws i ->
      if b land (1 lsl i) > 0 then ws + wa.(i) else ws) 0 in
    if ws <= lim then (
      let vs = range 0 t |> Seq.fold_left (fun vs i ->
        if b land (1 lsl i) > 0 then vs + va.(i) else vs) 0 in
      we.(b) <- ws;
      ve.(b) <- vs;
  ));
  (ve, we)

let solve x l =
  let module A = Array in
  let vs = A.of_list @@ items varr x in
  let ws = A.of_list @@ items warr x in
  let t = A.length vs in

  let lvs = A.sub vs 0 (t/2) in
  let rvs = A.sub vs (t/2) (t-t/2) in
  let lws = A.sub ws 0 (t/2) in
  let rws = A.sub ws (t/2) (t-t/2) in

  let ve1, we1 = enumerate (t/2) l lvs lws in
  let ve2, we2 = enumerate (t-t/2) l rvs rws in

  let ans = ref 0 in
  for i = 0 to (1 lsl (t/2)) - 1 do
    let v1 = ve1.(i) in
    let w1 = we1.(i) in
    for j = 0 to (1 lsl (t-t/2)) - 1 do
      let v2 = ve2.(j) in
      let w2 = we2.(j) in
      if w1 + w2 <= l then
        ans := max !ans (v1 + v2);
    done;
  done;
  !ans

let () =
  Scanf.scanf " %d" @@ fun q ->
  for i = 1 to q do
    Scanf.scanf " %d %d" solve
    |> Printf.printf "%d\n"
  done