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<bits/stdc++.h>
using namespace std;
const int N=1e5+7; bool fl[N];
int n,m,q,pos[N],que[N]; long long k,w[N],val[N];
inline long long read(){
	long long num=0; char t='+',g=getchar(); while(g<48||57<g) t=g,g=getchar();
	while(47<g&&g<58) num=(num<<1)+(num<<3)+g-48,g=getchar(); if(t=='-') return 0-num; return num;
}
inline void find(int u){
	que[++que[0]]=u,fl[u]=1; if(!fl[pos[u]]) find(pos[u]);
}
inline void getans(){
	int z=k%que[0]; for(int i=1;i<=que[0];i++) pos[que[i]]=que[(i+z-1)%que[0]+1];
}
int main(){
	n=read(); for(int i=1;i<=n;i++) w[i]=read(),pos[i]=i;
	m=read(),k=read(); for(int i=1;i<=m;i++) q=read(),swap(pos[q],pos[q+1]);
	for(int i=2;i<=n;i++) if(!fl[i]) que[0]=0,find(i),getans();
	for(int i=2;i<=n;i++) val[i]=w[i]-w[i-1]; long long vt=w[1]; printf("%lld\n",vt);
	for(int i=2;i<=n;i++) vt=vt+val[pos[i]],printf("%lld\n",vt);
	return 0;
}