Codeforces C. Destroying Array

C. Destroying Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array consisting of n non-negative integers a1, a2, ..., an.

You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.

After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

The third line contains a permutation of integers from 1 to n — the order used to destroy elements.

Output

Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.

Examples

input

Copy

4
1 3 2 5
3 4 1 2

output

Copy

5
4
3
0

input

Copy

5
1 2 3 4 5
4 2 3 5 1

output

Copy

6
5
5
1
0

input

Copy

8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6

output

Copy

18
16
11
8
8
6
6
0

Note

Consider the first sample:

1. Third element is destroyed. Array is now 1 3  *  5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3  *   * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now  *  3  *   * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.

题目大意：

由n个数值组成第一个序列，第二个序列的数字表示本次要消除第k个数值（隔开），输出每次消除后每段（即连续）序列的最大值。

解题思路：

直接暴力会超时。逆向思维，从最后消除的开始往前判断，用补充代替消除。每次补充一个数都判断其与前后数字是否连接，是的话用并查集连接起来，更新当前最大值。

#include
#include
#include
using namespace std;
const int N=100005; 
//大数据还是要开一个数组have来判断该数是否已经被访问，直接用visit判断会wa
int a[N],d[N],pre[N],have[N]; 
long long visit[N],ans[N];
int find(int x){
	if(x==pre[x])return x;
	return pre[x]=find(pre[x]);
}
int main(){
	int n,t=0;
	long long maxn=0;
	cin>>n;
	for(int i=1;i<=n;i++)pre[i]=i;
	for(int i=1;i<=n;i++)cin>>a[i];
	for(int i=1;i<=n;i++)cin>>d[i];
	for(int i=n;i>1;i--){
		int k=d[i];
		have[k]=1;
		long long num=a[k];
		visit[k]=a[k];
		if(have[k+1]){
			int r=find(k+1);
			pre[r]=k;
			num+=visit[r];
		}
		if(have[k-1]){
			int r=find(k-1);
			pre[r]=k;
			num+=visit[r];
		}
		visit[k]=num;
		maxn=max(num,maxn);
		ans[t++]=maxn;
	}
	for(int i=t-1;i>=0;i--){
		cout<