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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

//AUTHOR: RAVAN_2070
//PUNE INSTITUTE OF COMPUTER TECHNOLOGY
//Atcoder Educational DP
//N-Slimes
/*
    I ♥ CLARICE STARLING

    EXPLAINATION BELOW->
    Dp+segment tree

*/
//Iterative Solution
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pb push_back
#define all(x) x.begin(),x.end()
#define fo(i,a,b) for(i=a;i<b;i++)
#define foe(i,a,b) for(i=a;i<=b;i++)
#define MOD7 1000000007
const ll INF=(ll)1e18+5;
const int N=(int)2e5+5;
int n;
ll a[N],h[N];
ll dp[N],segment_tree[2*N+1];
/*void update(int node,int start,int end,int idx,ll val)
{
    if(start==end)segment_tree[node]=max(segment_tree[node],val);
    else
    {
        int mid=start+(end-start)/2;
        if(start>=idx&&idx<=mid)update(2*node,start,mid,idx,val);
        else update(2*node+1,mid+1,end,idx,val);
        
        segment_tree[node]=max(segment_tree[2*node],segment_tree[2*node+1]);
    }
}*/
void update(int node,int st,int en,int id,long long val)
{
    if(st>en || st>id || en<id) return ;
    if(st==en && st==id){
        segment_tree[node]=max(segment_tree[node],val);
        return ;
    }
    int mid=(st+en)/2;
    if(st<=id && id<=mid){
        update(2*node,st,mid,id,val);
    }
    else{
        update(2*node+1,mid+1,en,id,val);
    }	
    segment_tree[node]=max(segment_tree[2*node],segment_tree[2*node+1]);
}
ll query(int node,int start,int end,int l,int r)
{
    if(start>r||end<l)return 0;
    if(start>=l&&end<=r)return segment_tree[node];
    int mid=start+(end-start)/2;
    return max(query(2*node,start,mid,l,r),query(2*node+1,mid+1,end,l,r));
}
//dp[i] indicates max possible sum when i is the last height considered 
void solve()
{
    int i;cin>>n;
    foe(i,1,n)cin>>h[i];
    foe(i,1,n)cin>>a[i];
    dp[0]=0;
    
    ll ans=0;
    foe(i,1,n)
    {
       ll tmp=query(1,1,n,1,h[i]-1);
        dp[h[i]]=a[i]+tmp;
        update(1,1,n,h[i],dp[h[i]]);
    }
    
    foe(i,0,n)ans=max(ans,dp[i]);
    cout<<ans<<"\n";

}
int main()
{
   fastio
   solve();
  return 0;
}