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: Runtime Error)

/*
	$$$$$$$\                      $$\           $$$$$$$\
	$$  __$$\                     \__|          $$  __$$\
	$$ |  $$ | $$$$$$\   $$$$$$\  $$\  $$$$$$$\ $$ |  $$ | $$$$$$\   $$$$$$\   $$$$$$$\ $$$$$$\
	$$$$$$$\ |$$  __$$\ $$  __$$\ $$ |$$  _____|$$$$$$$\ | \____$$\ $$  __$$\ $$  _____|\____$$\
	$$  __$$\ $$ /  $$ |$$ |  \__|$$ |\$$$$$$\  $$  __$$\  $$$$$$$ |$$ |  \__|$$ /      $$$$$$$ |
	$$ |  $$ |$$ |  $$ |$$ |      $$ | \____$$\ $$ |  $$ |$$  __$$ |$$ |      $$ |     $$  __$$ |
	$$$$$$$  |\$$$$$$  |$$ |      $$ |$$$$$$$  |$$$$$$$  |\$$$$$$$ |$$ |      \$$$$$$$\\$$$$$$$ |
	\_______/  \______/ \__|      \__|\_______/ \_______/  \_______|\__|       \_______|\_______|
*/
#include <bits/stdc++.h>

using namespace std;

#define PB push_back
#define MP make_pair
#define INS insert
#define LB lower_bound
#define UB upper_bound
#define pii pair <int,int>
#define pll pair <long long, long long>
#define si pair<string, int>
#define is pair<int, string>
#define X first
#define Y second
#define _ << " " <<
#define sz(x) (int)x.size()
#define all(a) (a).begin(),(a).end()
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORD(i, a, b) for (int i = (a); i > (b); --i)
#define FORR(i, l, r) for (int i = (l); i <= (r); ++i)
#define FORP(i, a, b) for ((i) = (a); (i) < (b); ++i)
#define FORA(i, x) for (auto &i : x)
#define REP(i, n) FOR(i, 0, n)
#define BITS(x) __builtin_popcount(x)
#define MSET memset
#define MCPY memcpy
#define SQ(a) (a) * (a)

typedef long long ll;
typedef long double ld;
typedef vector <int> vi;
typedef vector <pii> vpi;
typedef vector <ll> vll;
typedef vector <pll> vpl;
typedef vector <double> vd;
typedef vector <ld> vld;
typedef vector<si> vsi;
typedef vector<is> vis;
typedef vector<string> vs;
//((float) t)/CLOCKS_PER_SEC

const int MOD = 1e9 + 7;
const int PI = acos(-1);
const int LOG = 20;
const int INF = 1e9 + 10;
const ll INFL = 1e18 + 10;
const int ABC = 30;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
const int dox[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int doy[] = {0, 0, -1, 1, -1, 1, -1, 1};

inline int sum(int a, int b){
  if (a + b >= MOD)
  	return a + b - MOD;
  if (a + b < 0)
  	return a + b + MOD;
  return a + b;
}

inline void add(int &a, int b){
  a = sum(a, b);
}

inline int mul(int a, int b){
  return (ll)a * (ll)b % MOD;
}

inline int sub(int a, int b){
	return (a - b + MOD) % MOD;
}

inline int pot(ll pot, int n){
	ll ret = 1;
	while (n){
		if (n & 1)
			ret = (ret * pot) % MOD;
		pot = (pot * pot) % MOD;
		n >>= 1;
	}
	return ret;
}

inline int divide(int a, int b){
	return mul(a, pot(b, MOD - 2));
}

ll lcm(ll a, ll b){
	return abs(a * b) / __gcd(a, b);
}

inline double ccw(pii A, pii B, pii C){
	return (A.X * B.Y) - (A.Y * B.X) + (B.X * C.Y) - (B.Y * C.X) + (C.X * A.Y) - (C.Y * A.X);
}

inline int CCW(pii A, pii B, pii C){
	double val = ccw(A, B, C);
	double eps = max(max(abs(A.X), abs(A.Y)), max(max(abs(B.X), abs(B.Y)), max(abs(C.X), abs(C.Y)))) / 1e9;
	if (val <= -eps)
		return -1;
	if (val >= eps)
		return 1;
	return 0;
}

void to_upper(string &x){
	REP(i, sz(x))
		x[i] = toupper(x[i]);
}

void to_lower(string &x){
	REP(i, sz(x))
		x[i] = tolower(x[i]);
}

string its(ll x){
	if (x == 0)
		return "0";
	string ret = "";
	while (x > 0){
		ret += (x % 10) + '0';
		x /= 10;
	}
	reverse(all(ret));
	return ret;
}

ll sti(string s){
	ll ret = 0;
	REP(i, sz(s)){
		ret *= 10;
		ret += (s[i] - '0');
	}
	return ret;	
}

const int N = 1e5 + 10;

int n, sor[N]; 
vll h, b;
vpl v;

struct Loga{
	vll f;
	int n;

	Loga(int _n){
		n = _n;
		f.resize(n, 0);
	}

	void update(int x, ll val){
		for (x++; x < n; x += (x & -x))
			f[x] = max(f[x], val);
	}
 
	ll query(int x){
		ll ret = 0;
		for (; x > 0; x -= (x & -x))
			ret = max(ret, f[x]);
		return ret;
	}

};

int main ()	{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	cin >> n;
	REP(i, n){
		ll x; cin >> x;
		h.PB(x);
		v.PB({x, i});
	}
	REP(i, n){
		ll x; cin >> x;
		b.PB(x);
	}

	sort(all(v));
	REP(i, sz(v))
		sor[v[i].Y] = i + 1;

	Loga dp(n + 10);
	REP(i, n){
		//cout << dp.query(sor[i]) << '\n';
		dp.update(sor[i], dp.query(sor[i]) + b[i]);
	}

	cout << dp.query(n + 1) << '\n';

	return 0;
}