[CF722C] 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, ..., a_n.

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, ..., a_n (0 ≤ a_i ≤ 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.

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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如果正着考虑我们不好处理删除元素；

所以时光倒流， 我们从后往前加元素， 这样题就变成了加入元素；

于是我们可以用并查集维护联通性， 每次加入一个元素， 我们就可以把它两边的联通块合并起来；

然后每次取max就是当前的答案；

有一个细节，就是你不能合并一个还未“出现”的元素， 所以我们要开一个bool数组判断是否出现过；

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#include <iostream>
#include <cstdio>
using namespace std;
int n, a[100005], p[100005];
int fa[100005];
long long val[100005];
int Find(int x){return x==fa[x]?x:fa[x]=Find(fa[x]);}
long long ans;
long long put[100005];
bool des[100005];
int main()
{
    scanf("%d", &n);
    for (register int i = 1 ; i <= n ; i ++) scanf("%d", &a[i]);
    for (register int i = 1 ; i <= n ; i ++) scanf("%d", &p[i]);
    for (register int i = 0 ; i <= n + 1 ; i ++) fa[i] = i;
    for (register int i = n ; i >= 1 ; i --){
        put[i] = ans;
        des[p[i]] = 1;
        int l = p[i] - 1, r = p[i] + 1;
        int fl = Find(l), fr = Find(r), fp = Find(p[i]);
        if (fl != fp and des[l])fa[fl] = fp, val[fp] += val[fl];
        if (fr != fp and des[r])fa[fr] = fp, val[fp] += val[fr];
        val[fp] += a[p[i]];
        ans = max(ans, val[fp]);
    }
    for (register int i = 1 ; i <= n ; i ++) printf("%lld\n", put[i]);
    return 0;
}