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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
import java.io.*;

public class Main {

	public static long[] tree;

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		int[] height = new int[n];
		int[] beauty = new int[n];
		long[] dp = new long[n + 1];
		String[] s = br.readLine().split(" ");
		for (int i = 0; i < n; i++)
			height[i] = Integer.parseInt(s[i]);

		s = br.readLine().split(" ");
		for (int i = 0; i < n; i++)
			beauty[i] = Integer.parseInt(s[i]);

		int b = 1;
		while (b < n)
			b *= 2;
		tree = new long[b * 2]; // segment tree stores max beauty in given range
								// we treat the segment tree as storing beauties of flowers w/
								// heights going from 1 -> n
		for (int i = 0; i < n; i++) {
			int k = b + height[i]-1;
			long max = 0;
			while (k > 1) { // because queries will always be from 1-height[i], we never care about right
								// side only care about left side (we are on right side)
//				System.out.println(k);
				if (k % 2 == 1)
					max = Math.max(max, tree[k - 1]); //bottom up; if sibling, compare
				k /= 2;
			}
//			System.out.println(max + "*");
			dp[height[i]] = beauty[i] + max; // dp[x] now represents the max beauty of flower w/
												// height i where it currently stands. if shorter flower
												// comes later, then shorter will have 0, which is ignored.
			k =b + height[i]-1;
			while (k >= 1) {
				tree[k] = Math.max(tree[k], dp[height[i]]);
				k /= 2;
			}
			
//			for (int e = 1; e < b*2; e++)
//				System.out.print(tree[e] + " ");
//			System.out.println();
//			for (int j = i; j >= 0; j--) {
//				if (height[j] < height[i] && dp[j] > max)
//					max = dp[j];
//			}
			// old n^2 approach TLE where dp[i] represents
			// the max beauty if the i'th flower was chosen

		}
		System.out.println(tree[1]);
//		for (long t : dp) System.out.print(t + " ");
//		System.out.println(dp[n - 1]);
	}

}