Problem:
Score : 100 points

Problem Statement

There are N flowers arranged in a row.
For each i (1 \leq i \leq N), the height and the beauty of the i-th flower from the left is h_i and a_i, respectively.
Here, h_1, h_2, \ldots, h_N are all distinct.

Taro is pulling out some flowers so that the following condition is met:

The heights of the remaining flowers are monotonically increasing from left to right.

Find the maximum possible sum of the beauties of the remaining flowers.

Constraints

All values in input are integers.

1 \leq N \leq 2 × 10^5

1 \leq h_i \leq N

h_1, h_2, \ldots, h_N are all distinct.

1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
h_1 h_2 \ldots h_N
a_1 a_2 \ldots a_N

Output

Print the maximum possible sum of the beauties of the remaining flowers.

Sample Input 1

4
3 1 4 2
10 20 30 40

Sample Output 1

60

We should keep the second and fourth flowers from the left.
Then, the heights would be 1, 2 from left to right, which is monotonically increasing, and the sum of the beauties would be 20 + 40 = 60.

Sample Input 2

1
1
10

Sample Output 2

10

The condition is met already at the beginning.

Sample Input 3

5
1 2 3 4 5
1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 3

5000000000

The answer may not fit into a 32-bit integer type.

Sample Input 4

9
4 2 5 8 3 6 1 7 9
6 8 8 4 6 3 5 7 5

Sample Output 4

31

We should keep the second, third, sixth, eighth and ninth flowers from the left.

Solution in Haskell (GHC 7.10.3) (judge verdict: Time Limit Exceeded)

import Data.List
import Data.Bits ((.&.))
import Data.Array.Unboxed
import Data.Array.MArray
import Data.Array.ST
import Control.Monad
import Control.Monad.ST

type Elem = Int
type BIT s = STUArray s Int Elem

{-
BITについて
sup2によりnを２の冪にする
配列をdp[1]~[n] 変数はi'
添え字iに対して、親ノードはi`div`2、子ノードは2*i,2*i+1
それぞれPは一点、Rは範囲の操作
fが群の演算ならば（逆元が存在すれば）、任意の範囲を取り出すことが可能
モノイドならば(1,n)のような範囲を取り出し可能
-}

--nを超える最小の２の冪
sup2 :: Int -> Int
sup2 n = 2^(ceiling $ logBase 2 (fromIntegral n :: Double) :: Int)

newBIT :: MArray a e m => Int -> e -> m (a Int e)
newBIT n e = newArray (1,n) e

--logN
readRBIT' :: BIT s -> (Elem -> Elem -> Elem) -> Int -> Int -> ST s Elem
readRBIT' bit f i partial
  | 1 > i = return partial
  | otherwise = do
    a <- readArray bit i
    readRBIT' bit f (i-(i .&. (-i))) (f a partial)

readPBIT :: BIT s -> (Elem -> Elem -> Elem) -> Int -> Int -> ST s Elem
readPBIT bit f i partial
  = do
    (_,a) <- getBounds bit
    if a < i then return partial
      else do
        a <- readArray bit i
        readPBIT bit f (i+(i .&. (-i))) (f a partial)

--(1,i)の範囲で更新
modifyRBIT' :: BIT s -> (Elem -> Elem) -> Int -> ST s ()
modifyRBIT' bit f i
  | i < 1 = return ()
  | otherwise  = do
   readArray bit i >>= writeArray bit i .f
   modifyRBIT' bit f (i-(i .&. (-i)))

modifyPBIT :: BIT s -> (Elem -> Elem) -> Int -> ST s ()
modifyPBIT bit f i
   = do
   (_,a) <- getBounds bit
   if a < i then return ()
     else do
        now <- readArray bit i
        writeArray bit i $ f now
        --traceShowM (i,now,f now)
        modifyPBIT bit f (i+(i .&. (-i)))

solve :: Int -> [Int] -> [Int] -> ST s Elem
solve n hs as = do
  bit' <- newBIT n 0
  solve' bit' n hs as
  where
    solve' bit n [] _ = readPBIT bit max n 0
    solve' bit n (h:hs) (a:as) = do
      a' <- readRBIT' bit max h 0
      modifyPBIT bit (max $ a+a') h
      solve' bit n hs as

main :: IO ()
main = do
  n <- readLn
  hs <- map read.words <$> getLine
  as <- map read.words <$> getLine
  print $ runST $ solve n hs as