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<cstring>
using namespace std;
const int N=5e5;
int x[N],opt[N],id[N],p[N],vis[N],n,m,tmp[N];
long long s[N],K;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",x+i),id[i]=i;
    for(int i=n;i>=1;i--)x[i]-=x[i-1];
    scanf("%d%lld",&m,&K);
    for(int i=1;i<=m;i++)scanf("%d",opt+i),swap(id[opt[i]],id[opt[i]+1]);
    for(int i=1;i<=n;i++)p[id[i]]=i;
    for(int i=1;i<=n;i++)
    {
        if(vis[i])continue;
        int u=i,len=0;tmp[0]=u;vis[u]=1;
        while(p[u]!=i)u=p[u],tmp[++len]=u,vis[u]=1;
        ++len;
        for(int i=0;i<len;i++)p[tmp[i]]=tmp[(i+K)%len];
    }
    for(int i=1;i<=n;i++)id[p[i]]=i;
    
    for(int i=1;i<=n;i++)s[i]=s[i-1]+x[id[i]],printf("%.1lf\n",(double)s[i]);
}