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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define int long long
#define seg_tree_valtype int
#define inf (int)(1e18)

typedef struct {
	seg_tree_valtype *dat;
	int n; //要素数
	int N; //要素数以上の最小の2のべき乗
	seg_tree_valtype (*f)(seg_tree_valtype x, seg_tree_valtype y);//演算
	seg_tree_valtype e;//演算の単位元
}seg_tree;

//要素n, 演算f, 単位元eのsegment_treeを作る
seg_tree *make_seg_tree(int n, seg_tree_valtype (*f)(seg_tree_valtype x, seg_tree_valtype y), seg_tree_valtype e){
	seg_tree *t = (seg_tree *)malloc(sizeof(seg_tree));
	int N, i;
	for(N = 1; N < n; N *= 2){}
	t->dat = (seg_tree_valtype *)malloc(sizeof(seg_tree_valtype) * (2 * N - 1));
	for(i = 0; i < 2 * N - 1; i++){
		t->dat[i] = e;
	}
	t->n = n;
	t->N = N;
	t->f = f;
	t->e = e;
	return t;
}

//i番目の要素を返す
seg_tree_valtype look_val(int i, seg_tree *t){
	if(i < 0 || t->n <= i){
		printf("In function 'look_val':\nrefered out of array\n");
	}
	return t->dat[i + t->N - 1];
}

//i番目の値をxに更新する
void update(int i, seg_tree_valtype val, seg_tree *t){
	if(i < 0 || t->n <= i){
		printf("In function 'update':\nrefered out of array\n");
	}
	i += t->N - 1;
	t->dat[i] = val;
	while(i > 0){
		i = (i - 1) / 2;
		t->dat[i] = (t->f)(t->dat[2 * i + 1], t->dat[2 * i + 2]);
	}
}

seg_tree_valtype query_sub(int a, int b, int k, int l, int r, seg_tree *t){
	if(r <= a || b <= l){
		return t->e;
	}
	else if(a <= l && r <= b){
		return t->dat[k];
	}
	else{
		int h = (l + r) / 2;
		seg_tree_valtype vl = query_sub(a, b, 2 * k + 1, l, h, t);
		seg_tree_valtype vr = query_sub(a, b, 2 * k + 2, h, r, t);
		return (t->f)(vl, vr);
	}
}

//[a, b)の区間に対しての演算の結果を返す
seg_tree_valtype query(int a, int b, seg_tree *t){
	if(a < 0 || b < 0 || t->n < a || t->n < b){
		printf("In function 'query':\nrefered out of array\n");
	}
	if(a >= b){
		return t->e;
	}
	else{
		return query_sub(a, b, 0, 0, t->N, t);
	}
}

int max(int a, int b){
	if(a > b){
		return a;
	}
	else{
		return b;
	}
}

signed main(){
	int N, i;
	scanf("%lld", &N);
	int *h = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		scanf("%lld", &h[i]);
	}
	int *a = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		scanf("%lld", &a[i]);
	}
	seg_tree *t = make_seg_tree(N + 1, max, -inf);
	update(0, 0, t);
	for(i = 0; i < N; i++){
		update(h[i], query(0, h[i], t) + a[i], t);
	}
	printf("%lld\n", query(0, N + 1, t));
	return 0;
}