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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
import java.text.*;
//Solution Credits: Taranpreet Singh
public class Main{
    //SOLUTION BEGIN
    void pre() throws Exception{}
    void solve(int TC) throws Exception{
        int n = ni();
        SegTree t = new SegTree(1+n);
        int[] h = new int[n];
        for(int i = 0; i< n; i++)h[i] = ni();
        for(int i = 0; i< n; i++){
            long a = nl();
            t.update(h[i], t.mx(0,h[i]-1)+a);
        }
        pn(t.t[1]);
    }
    class SegTree{
        int m = 1;
        long[] t;
        public SegTree(int n){
            while(m<n)m<<=1;
            t= new long[m<<1];
        }
        void update(int p, long v){
            t[p+=m] = v;
            for(p>>=1; p>0; p>>=1)t[p] = Math.max(t[p<<1], t[p<<1|1]);
        }
        long mx(int l, int r){
            return mx(l,r,0,m-1,1);
        }
        long mx(int l, int r, int ll, int rr, int i){
            if(l==ll && r==rr)return t[i];
            int mid = (ll+rr)>>1;
            if(r<=mid)return mx(l,r,ll,mid,i<<1);
            else if(l>mid)return mx(l,r,mid+1,rr,i<<1|1);
            else return Math.max(mx(l,mid,ll,mid,i<<1), mx(mid+1, r, mid+1, rr, i<<1|1));
        }
    }
    //SOLUTION END
    void hold()throws Exception{throw new Exception("Hold right there, Sparky!");}
    long mod = (long)1e9+7, IINF = (long)4e18;
    final int INF = (int)2e9, MX = (int)1e6+1;
    DecimalFormat df = new DecimalFormat("0.000000000000000");
    double PI = 3.1415926535897932384626433832792884197169399375105820974944, eps = 1e-8;
    static boolean multipleTC = false, memory = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        int T = (multipleTC)?ni():1;
        //Solution Credits: Taranpreet Singh
        pre();for(int i = 1; i<= T; i++)solve(i);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n(){return in.next();}
    String nln(){return in.nextLine();}
    int ni(){return Integer.parseInt(in.next());}
    long nl(){return Long.parseLong(in.next());}
    double nd(){return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }  
            return str;
        }
    }
}