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.
題意：按著第二個(gè)數(shù)組給出的數(shù)組序列刪除元素，每一次刪除之后問(wèn)最大的連續(xù)序列的和是多少。
思路：刪除的不好弄，那么我們就倒著來(lái)，按著增加的來(lái)算。對(duì)于位置i的元素，我們嘗試著去把它和i+1和i-1合并，優(yōu)先隊(duì)列維護(hù)最大值。
代碼如下：

#include<bits/stdc++.h> #define ll long long using 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?yàn)閿?shù)組元素有可能為0，所以起始值設(shè)置為-1。ans[n]=0;priority_queue<ll> 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<<ans[i]<<endl; }

努力加油a啊，(^o)/~

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