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: Wrong Answer)

open Batteries
let n = read_int ()
let dp = Array.init n @@ fun _ -> [|0; 0|]
let rec vw_loop i =
  if i >= n then () else (
    Scanf.sscanf (read_line()) "%d %d" (fun v w -> 
      dp.(i) <- [|v; w|]
    ); vw_loop (i+1)
  )
let _ = vw_loop 0
let dp2 = Array.init (n+1) @@ fun _ -> []
let rec dp_loop i =
  if i >= n then () else (
    if i = 0 then (dp2.(i) <- dp.(i) :: dp2.(i); dp_loop (i+1)) else (
      dp2.(i) <- dp.(i) :: dp2.(i / 2); dp_loop (i+1)
    )
  )
let _ = dp_loop 0
let q = read_int ()

(* ここからナップザック *)
let nap items l =
(* 価値；重さ*)
  let n_ = Array.length items in
  let nap_dp = Array.init (l+1) @@ fun _ -> 0 in
  for i = 0 to (n_-1) do
    let rec w_loop w =
      if w < 0 then () else (
        if w - items.(i).(1) >= 0 then (
          nap_dp.(w) <- max (nap_dp.(w)) (nap_dp.(w - items.(i).(1)) + items.(i).(0));
          w_loop (w-1)
        ) else w_loop (w-1)
      ) in w_loop l
  done; nap_dp.(l)
  

(* nap [|[|3;2|];[|2;1|];[|6;3|];[|1;2|];[|3;1|];[|85;5|]|] 9;; *)

let rec q_loop i =
  if i >= q then () else (
    Scanf.sscanf (read_line()) "%d %d" (fun v l -> 
      Printf.printf "%d\n" @@ nap (Array.of_list dp2.(v-1)) l
    ); q_loop (i+1)
  )

let _ = q_loop 0