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: Accepted)

let () = Scanf.scanf "%d\n" @@ fun n ->
  let vs = Array.make n 0 in
  let ws = Array.make n 0 in
  for i = 0 to n - 1 do
    Scanf.scanf "%d %d\n" @@ fun v w ->
      vs.(i) <- v;
      ws.(i) <- w
  done;
  Scanf.scanf "%d\n" @@ fun q ->
    let vls = Array.init q @@ fun _ -> Scanf.scanf "%d %d\n" @@ fun v l -> v, l in
    let maxl = Array.fold_right (fun (_, l) -> max l) vls min_int in
    let n' = min n ((1 lsl 10) - 1) in
    let dp = Bigarray.Array2.create Bigarray.int Bigarray.c_layout (n' + 1) (maxl + 1) in
    Bigarray.Array2.fill (Bigarray.Array2.sub_left dp 0 1) 0;
    for i = 1 to n' do
      let v = Array.unsafe_get vs (i - 1) in
      let w = Array.unsafe_get ws (i - 1) in
      Bigarray.Array2.blit
        (Bigarray.Array2.sub_left dp (i lsr 1) 1)
        (Bigarray.Array2.sub_left dp i 1);
      for j = w to maxl do
        let d = Bigarray.Array2.unsafe_get dp (i lsr 1) j in
        let d' = v + Bigarray.Array2.unsafe_get dp (i lsr 1) (j - w) in
        if d < d' then Bigarray.Array2.unsafe_set dp i j d'
      done
    done;
    let rec solve l i s m =
      if l < 0
      then m
      else if i <= n'
      then max m (s + Bigarray.Array2.unsafe_get dp i l)
      else solve l (i lsr 1) s @@ solve (l - Array.unsafe_get ws (i - 1)) (i lsr 1) (s + Array.unsafe_get vs (i - 1)) m in
    Array.iter (fun (v, l) -> Printf.printf "%d\n" @@ solve l v 0 min_int) vls