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.Array  --変数衝突
import Data.Array.Unboxed
import Data.Array.MArray
import Data.Array.ST
import qualified Data.Vector.Unboxed as V (last,(!),fromList,Vector)
import Control.Monad
import Control.Monad.ST

type DP s = STUArray s Int Int

{-
セグ木について
sup2によりnを２の冪にする
配列をdp[1]~[n] 変数はi'
セグ木をsegT[1]~[2*n-1] 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
高さは (floor logBase 2 (fromIntegral i))+1
関数名についている1は葉を表している
-}
-- n+n/2+n/4+...<=2*n-1だが、子ノードの定義の仕方により2*n-1を超えることがある。

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

newSegT :: MArray a e m => (Int,Int) -> e -> m (a Int e)
newSegT (1,n) e = newArray (1,2*n-1) e

read1SegT :: MArray a e m => a Int e -> Int -> Int -> m e
read1SegT dparr i' i = readArray dparr (i'+i-1)

write1SegT :: MArray a e m => a Int e -> Int -> Int -> e -> m ()
write1SegT dparr i' i e = writeArray dparr (i'+i-1) e

--logN
readPSegT :: DP s -> (Int,Int) -> Int -> Int -> Int -> Int -> ST s Int
readPSegT dparr (l,r) n i' i max0
  | 2*n-1 < i = return max0 ---Error??-----
  | otherwise = do
    let mid = (l+r)`div`2
    a <- readArray dparr i
    if mid < i'
      then readPSegT dparr (mid+1,r) n i' (2*i+1) (max a max0)
      else readPSegT dparr (l,mid) n i' (2*i) (max a max0)

--セグ木の更新,O((logN)^2)
--(left,right)の範囲で更新
modifyRSegT :: DP s -> (Int,Int) -> (Int,Int) -> Int -> Int -> Int -> ST s ()
modifyRSegT dparr (left,right) (l,r) a n i
  | right < left = return ()
  | 2*n-1 < i   = error $ "modifyRSegT Error : i = " ++ (show i) ++ (show ((left,right), (l,r), a)) ---Error??
  | left == l && right >= r = do
   a' <- readArray dparr i
   writeArray dparr i (max a' a)
   modifyRSegT dparr (r+1,right) (1,n) a n 1
  | otherwise  = do
   let mid = (l+r)`div`2
   if mid < left
     then modifyRSegT dparr (left,right) (mid+1,r) a n (2*i+1)
     else modifyRSegT dparr (left,right) (l,mid) a n (2*i)

dp_i :: DP s -> Int -> [Int] -> [Int] -> ST s ()
dp_i dparr n [] [] = return ()
dp_i dparr n (h:hs) (a:as) = do
  nowa <- readPSegT dparr (1,n) n h 1 0
  modifyRSegT dparr (h,n) (1,n) (nowa+a) n 1
  dp_i dparr n hs as

solve :: Int -> [Int] -> [Int] -> ST s Int
solve n hs as = do
  dparr <- newSegT (1,n) 0
  dp_i dparr n hs as
  readPSegT dparr (1,n) n n 1 0

main :: IO ()
main = do
 n  <- read <$> getLine
 hs <- map read.words <$> getLine
 as <- map read.words <$> getLine
 putStrLn $ show $ runST $ solve (sup2 n) hs as