codefoces 722C - 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

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.

题意：每次破坏一个数，使得剩下的联通块和最大。。。

仔细想想，这是并查集的相反操作。。每次破坏看成每次加入一个数，并且旁边两个合成一个联通块，n次操作只有，就是倒着操作了。

#include
using namespace std;
typedef long long ll;
const int N=1e5+100;
int f[N];
void init()
{
    for(int i=0; i<=N; i++)
        f[i]=i;
}
int a[N],b[N],vis[N];
ll sum[N];
vectorans;
int find(int x)
{
    return f[x]==x?x:f[x]=find(f[x]);
}
void hebing(int x,int y)
{
    f[x]=y;
    sum[y]+=sum[x];
}
int main()
{
    int n;
    scanf("%d",&n);
    init();
    memset(vis,0,sizeof(vis));
    memset(sum,0,sizeof(sum));
    for(int i=1; i<=n; i++)
        scanf("%d",&a[i]);
    for(int i=1; i<=n; i++)
        scanf("%d",&b[i]);
    ll res=0;
    ans.clear();
    for(int i=n; i>=1; i--)
    {
        ans.push_back(res);
        int x=b[i];
        vis[x]=1;
        sum[x]=a[x];
        if(vis[x+1])
        {
            int y=find(x+1);
            hebing(y,x);
        }
        if(vis[x-1])
        {
            int y=find(x-1);
            hebing(y,x);
        }
        res=max(res,sum[x]);
    }
    int sz=ans.size();
    for(int i=sz-1; i>=0; i--)
        cout<