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 maxn 100100
using namespace std;
typedef long long ll;
int n,m,a[maxn],b[maxn],c[maxn],d[maxn];
ll x[maxn];
ll k;
void mul(int a[],int b[],int c[]){
//	for(int i=1;i<=n;++i)printf("[%d]",a[i]);puts("");
//	for(int i=1;i<=n;++i)printf("[%d]",b[i]);puts("");
	for(int i=1;i<=n;++i)d[i]=a[b[i]];
	for(int i=1;i<=n;++i)c[i]=d[i];
//	for(int i=1;i<=n;++i)printf("[%d]",c[i]);puts("->");
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)scanf("%lld",&x[i]);
	for(int i=n;i>=1;--i)x[i]-=x[i-1];
	scanf("%d%lld",&m,&k);
	for(int j=1;j<=m;++j)scanf("%d",&a[j]);
	for(int j=1;j<=n;++j)c[j]=b[j]=j;
	for(int j=1;j<=m;++j)swap(b[a[j]],b[a[j]+1]);
	for(;k;k>>=1,mul(b,b,b))
		if(k&1)mul(c,b,c);
	ll tg=0;
	for(int i=1;i<=n;++i)
		tg=(tg+x[c[i]]),printf("%lld\n",tg);
}