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 Haskell (GHC 8.8.3) (judge verdict: Time Limit Exceeded)

import qualified Data.ByteString.Char8 as BS
import Data.Char
import Data.List
import Control.Monad
import Data.Array
import qualified Data.IntMap as M

main = do
  li <- BS.getLine
  let Just (n,_) = BS.readInt li
  vws <- forM [1..n] get2
  li <- BS.getLine
  let Just (q,_) = BS.readInt li
  vls <- forM [1..q] get2
  let ans = compute n vws vls
  mapM_ print ans

get2 _ = do
  li <- BS.getLine
  let (a:b:_) = unfoldr (BS.readInt . BS.dropWhile isSpace) li
  return (a,b)

compute :: Int -> [(Int,Int)] -> [(Int,Int)] -> [Int]
compute n vws vls =
    [ fst $ M.findMax $ M.filter (l >=) (nstr ! v) | (v,l) <- vls ]
  where
    nstr = listArray (0,n) nslist
    nslist = M.fromList [(0,0)] :
      [ M.unionWith min ns $ M.mapKeys (v +) $ M.map (w +) ns
      | (i,(v,w)) <- zip [1..n] vws
      , let ns = nstr ! (div i 2)
      ]
{-
根から各頂点の方向に、総当たりの価値の組み合わせを作っておき、上限で切る。
ただし遅延評価なので必要なところしか計算しないという作戦。
これarray必要かな？後でindexでアクセスされるからという理由でそうするべきだね。
組み合わせは、価値をkey、それをつくる最も軽い重量をvalにしたIntMapで管理しよう。
-}