CF722C. Destroying Array[并查集 离线]

链接：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

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: 

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.

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题意：正整数序列，每次毁灭一个元素，每次求剩下元素不包含毁灭元素的线段的最大和

 

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yy了各种方法，BIT，线段树，甚至各种STL，然而.....

 

删除不好处理，可以逆序加入点，用并查集维护区间连通

和JSOI2008星球大战一个想法，倒着处理

如果i根i+1连通，那么sum[i]+=sum[find(i+1)]，再把find(i+1)合并到i里（注意必须是这个合并顺序）

i-1同理

//
//  main.cpp
//  c
//
//  Created by Candy on 10/1/16.
//  Copyright © 2016 Candy. All rights reserved.
//
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
const int N=1e5+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x;
}
int n,a[N],p[N],vis[N];
ll sum[N],mx=0,st[N],top=0;
int fa[N];
inline int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}

int main(){
    n=read();
    for(int i=1;i<=n;i++) a[i]=read(),fa[i]=i;
    for(int i=1;i<=n;i++) p[i]=read();
    for(int i=n;i>=1;i--){
        st[++top]=mx;
        int cur=p[i],f1=0,f2=0;
        vis[cur]=1; sum[cur]+=a[cur];
        if(vis[cur-1]){f1=find(cur-1);sum[cur]+=sum[f1];fa[f1]=cur;}
        if(vis[cur+1]){f2=find(cur+1);sum[cur]+=sum[f2];fa[f2]=find(cur);}
        mx=max(mx,sum[cur]);
    }
    while(top) printf("%I64d\n",st[top--]);
}