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)

// STANDARD LIS O(N*N)


// fahadmd16(IIEST Shibpur)
#include<bits/stdc++.h>
using namespace std;

#define ll                                      long long
#define pb                                      push_back
#define pf                                      push_front
#define eb                                      emplace_back
#define mp                                      make_pair
#define ff                                      first
#define ss                                      second
#define db                                      double
#define pll                                     pair<ll, ll>
#define ALL(a)                                  (a).begin(),(a).end()
#define endl                                    "\n"

#define forn(i, n) for (ll i = 0; i < n; i++)
#define forn1(i, n) for (ll i = 1; i < n; i++)
#define ford(i, n) for (ll i = n - 1; i >= 0; i--)
#define ford1(i, n) for (ll i = n - 1; i >= 1; i--)
#define forab(i, a, b) for(ll i = a; i < b; i++)
#define forabd(i, a, b) for(ll i = a; i >= b; i--)
#define eps 1e-10

#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)


#define IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

#define trace1(x)                cout<<#x<<": "<<x<<endl
#define trace2(x, y)             cout<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define trace3(x, y, z)          cout<<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define trace4(a, b, c, d)       cout<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl
#define trace5(a, b, c, d, e)    cout<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<endl
#define trace6(a, b, c, d, e, f) cout<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<" | "<<#f<<": "<<f<<endl
#define tracen( x , n ){ cout<<(#x)<<":";for(int i=0;i<n;i++)cout<<(#x)<<"["<<i<<"]:"<< x[i] << "    "; cout<<endl;}
#define tracenm( x , n , m ){ cout<<(#x)<<endl;for(int i=0;i<n;i++){ cout<<"Row #"<<i<<":"; for(int j=0;j<m;j++) cout<<x[i][j]<<"     "; cout << endl;}}
#define trace(v){ cout<<#v<<": ";for(auto it=v.begin();it!=v.end();it++)cout<<*it<<" ";cout<<endl;}
using namespace std;



const ll sz = 2e5 + 2;
const ll mod = 1e9 + 7;
const db PI = acos(-1);



typedef vector<ll> VL;
typedef vector<pll > VLL;
typedef vector<vector<ll> > VVL;



ll power(ll n, ll k, ll m) {
    ll ans = 1;
    while (k) {
        if (k & 1)ans = (ans * n) % m;
        k /= 2;
        n = (n * n) % m;
    }
    return ans;
}

ll tree[4 * sz + 5];
map<ll, ll> mpx;

void build(ll l, ll r, ll Tidx) {
    if (l == r) {
        tree[Tidx] = mpx[l];
        return;
    }
    ll mid = (l + r) / 2;
    build(l, mid, Tidx * 2);
    build(mid + 1, r, Tidx * 2 + 1);
    tree[Tidx] = max(tree[Tidx * 2], tree[Tidx * 2 + 1]);
}

void update(ll l, ll r, ll idx1, ll Tidx, ll val) {
    if (l == r) {
        tree[Tidx] = val;
        return;
    }

    ll mid = (l + r) / 2;
    if (idx1 <= mid)
        update(l, mid, idx1, Tidx * 2, val);
    else
        update(mid + 1, r, idx1, Tidx * 2 + 1, val);

    tree[Tidx] = max(tree[Tidx * 2], tree[Tidx * 2 + 1]);

}

ll query(ll l, ll r, ll Tidx, ll st, ll en) {
    if (r < st || l > en)return 0; // NO OVERLAP

    if (l >= st && r <= en) { // COMPLETE OVERLAP

        return tree[Tidx];
    }

    ll mid = (l + r) / 2;

    ll ans1 = query(l, mid, Tidx * 2, st, en);
    ll ans2 = query(mid + 1, r, Tidx * 2 + 1, st, en);

    return max(ans1, ans2);



}





void solver() {
    ll n; cin >> n;
    VLL a(n);
    forn(i, n)cin >> a[i].ff;
    forn(i, n) {cin >> a[i].ss; mpx[a[i].ff] = a[i].ss;}

    // build(1, n, 1);
    // update(1, n, 1, 1, 75);
    // cout << query(1, n, 1, 1, 1) << endl;


    forn1(i, 4 * n + 5) tree[i] = 0; //cout << tree[i] << " "; cout << endl;
    ll ans = 0;
    forn(i, n) {
        ll xx = query(1, n, 1, 1, a[i].ff);
        update(1, n, a[i].ff, 1, xx + a[i].ss);
        ans = max(ans, xx + a[i].ss);
    }
    cout << ans << endl;

    // tracen(dp, n);

}


int main() {

    IOS;
    ll T = 1; // cin >> T;
    forn1(i, T + 1) {
        clock_t z = clock();
        solver();
        debug("Total Time: %.3f\n", (double)(clock() - z) / CLOCKS_PER_SEC);
    }
    return 0;
}