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 <cstring>
#include <algorithm>
#include <cstdlib>
#include <queue>
using namespace std;
typedef long long ll;
#define N 100050
int n,a[N],m,p[N],vis[N],b[N],lb;ll X[N],d[N],K,ans[N];
int main() {
	scanf("%d",&n);
	int i,y=0,j;
	for(i=1;i<=n;i++) scanf("%lld",&X[i]),d[i]=X[i]-X[i-1];
	scanf("%d%lld",&m,&K);
	for(i=1;i<=m;i++) scanf("%d",&a[i]);
	for(i=1;i<=n;i++) p[i]=i;
	for(i=1;i<=m;i++) swap(p[a[i]],p[a[i]+1]);
	for(i=1;i<=n;i++) if(!vis[i]) {
		int x=i;lb=0;
		for(;!vis[x];x=p[x]) {
			vis[x]=1;
			b[lb++]=x;
		}
		for(int i=0;i<lb;i++) {
			ans[b[i]]=d[b[(i+K)%lb]];
		}
	}
	for(i=1;i<=n;i++) X[i]=ans[i]+X[i-1];
	for(i=1;i<=n;i++) printf("%lld.0\n",X[i]);
}
