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>
#define N 500005
#define int long long
using namespace std;
int n,m,xx,b[N],x[N],tmp[N],ans[N],sum[N],tmpp[N];long long K;
inline void change1(){
	for (int i=1;i<=n;i++) tmpp[i]=ans[i];
	for (int i=1;i<=n;i++) ans[i]=tmp[tmpp[i]];
}
inline void change2(){
	for (int i=1;i<=n;i++) tmpp[i]=tmp[i];
	for (int i=1;i<=n;i++) tmp[i]=tmpp[tmpp[i]];
}
signed main(){
	scanf("%lld",&n);
	for (int i=1;i<=n;i++) scanf("%lld",&x[i]),b[i]=x[i]-x[i-1];
	scanf("%lld%lld",&m,&K);
	for (int i=1;i<=n;i++) tmp[i]=i;
	for (int i=1;i<=m;i++){
		scanf("%lld",&xx);
		swap(tmp[xx],tmp[xx+1]);
	}
	for (int i=1;i<=n;i++) ans[i]=i;
	while (K){
		if (K&(int)1) change1();
		K=K/(int)2;change2();
	}
	for (int i=1;i<=n;i++) sum[i]=b[ans[i]];
	for (int i=1;i<=n;i++) sum[i]+=sum[i-1];
	for (int i=1;i<=n;i++) printf("%lld\n",sum[i]);
	return 0;
}