There are N rabbits on a number line. The rabbits are conveniently numbered 1 through N. The coordinate of the initial position of rabbit i is x_i.

The rabbits will now take exercise on the number line, by performing sets described below. A set consists of M jumps. The j-th jump of a set is performed by rabbit a_j (2≤a_j≤N-1). For this jump, either rabbit a_j-1 or rabbit a_j+1 is chosen with equal probability (let the chosen rabbit be rabbit x), then rabbit a_j will jump to the symmetric point of its current position with respect to rabbit x.

The rabbits will perform K sets in succession. For each rabbit, find the expected value of the coordinate of its eventual position after K sets are performed.

Constraints

* 3≤N≤10^5
* x_i is an integer.
* |x_i|≤10^9
* 1≤M≤10^5
* 2≤a_j≤N-1
* 1≤K≤10^{18}

Input

The input is given from Standard Input in the following format:


N
x_1 x_2 ... x_N
M K
a_1 a_2 ... a_M


Output

Print N lines. The i-th line should contain the expected value of the coordinate of the eventual position of rabbit i after K sets are performed. The output is considered correct if the absolute or relative error is at most 10^{-9}.

Examples

Input

3
-1 0 2
1 1
2


Output

-1.0
1.0
2.0


Input

3
1 -1 1
2 2
2 2


Output

1.0
-1.0
1.0


Input

5
0 1 3 6 10
3 10
2 3 4


Output

0.0
3.0
7.0
8.0
10.0
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
#define llong long long
int n,m;
int pos[200200];
int nxt[200200];
llong K;

bool vis[200200];
int sta[200200],top;

double ans[100100];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;++i) scanf("%d",&pos[i]),nxt[i]=i;
    for(int i=n;i;--i)pos[i]-=pos[i-1];
    scanf("%d%lld",&m,&K);
    int xx;
    for(int i=1;i<=m;++i)
        scanf("%d",&xx),swap(nxt[xx],nxt[xx+1]);
    for(int i=1;i<=n;++i)
        if(!vis[i])
        {
            xx=i; top=0;
            while(!vis[xx]) vis[xx]=1,sta[top++]=xx,xx=nxt[xx];
            for(int j=0;j<top;++j)
                ans[sta[j]]=pos[sta[(K+j)%top]];
        }
    for(int i=1;i<=n;++i)
    {
        ans[i]+=ans[i-1];
        printf("%.1lf ",ans[i]);
    }
    puts("");
    return 0;
}
