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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 4 1 3 2 5 3 4 1 2 Output 5 4 3 0 Input 5 1 2 3 4 5 4 2 3 5 1 Output 6 5 5 1 0 Input 8 5 5 4 4 6 6 5 5 5 2 8 7 1 3 4 6 Output 18 16 11 8 8 6 6 0 Note Consider the first sample:

Third element is destroyed. Array is now 1 3  *  5. Segment with maximum sum 5 consists of one integer 5.

Fourth element is destroyed. Array is now 1 3  *   * . Segment with maximum sum 4 consists of two integers 1 3. First element is destroyed. Array is now  *  3  *   * . Segment with maximum sum 3 consists of one integer 3. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0. 题意：按着第二个数组给出的数组序列删除元素，每一次删除之后问最大的连续序列的和是多少。 思路：删除的不好弄，那么我们就倒着来，按着增加的来算。对于位置i的元素，我们尝试着去把它和i+1和i-1合并，优先队列维护最大值。 代码如下：

#include
   
    #define ll long longusing namespace std;const int maxx=1e5+100;ll a[maxx];ll ans[maxx];int b[maxx];int f[maxx];ll dis[maxx];int n;inline int getf(int u){
   	return (u==f[u]?u:f[u]=getf(f[u]));}int main(){
   	scanf("%d",&n);	for(int i=1;i<=n;i++) scanf("%I64d",&a[i]);	for(int i=1;i<=n;i++) scanf("%d",&b[i]);	for(int i=1;i<=n;i++) f[i]=i,dis[i]=-1;//因为数组元素有可能为0，所以起始值设置为-1。	ans[n]=0;	priority_queue
    
      q;	q.push(0);	for(int j=n;j>=1;j--)	{
   		dis[b[j]]=a[b[j]];		q.push(dis[b[j]]);		int i=b[j];		if(i+1<=n&&dis[i+1]!=-1)		{
   			int t1=getf(i+1);			int t2=getf(i);			f[t1]=t2;			dis[t2]+=dis[t1];			dis[t1]=0;			q.push(dis[t2]);		}		if(i-1>=1&&dis[i-1]!=-1)		{
   			int t1=getf(i-1);			int t2=getf(i);			f[t1]=t2;			dis[t2]+=dis[t1];			dis[t1]=0;			q.push(dis[t2]);		}		ans[j-1]=q.top();	}	for(int i=1;i<=n;i++) cout<
     
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