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<cstdio>
#include<cctype>
const int maxn=100010;
int n,m;
long long k;
int d[maxn],p[maxn];
long long r[maxn];
bool vs[maxn];
int st[maxn],tp;
long long In() {
	register char c;
	for(;c=getchar(),c!='-'&&!isdigit(c););
	register bool f=c=='-';
	register long long s=f?0:c-'0';
	for(;c=getchar(),isdigit(c);)s=s*10+c-'0';
	return f?-s:s;
}
int main() {
	n=In();
	for(register int i=1;i<=n;++i)d[i]=In();
	for(register int i=n;i;--i)d[i]-=d[i-1];
	m=In(),k=In();
	for(register int i=1;i<=n;++i)p[i]=i;
	for(register int i=1,x;i<=m;++i)x=In(),p[x+1]^=p[x],p[x]^=p[x+1],p[x+1]^=p[x];
	for(register int i=1;i<=n;++i)if(!vs[i]) {
		tp=0;
		for(register int x=i;!vs[x];x=p[x])vs[x]=1,st[++tp]=x;
		for(register int j=1;j<=tp;++j)r[st[j]]=d[st[(j+k-1)%tp+1]];
	}
	for(register int i=1;i<=n;++i)printf("%lld.0\n",r[i]+=r[i-1]);
	return 0;
}