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 <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 100005;
int a[MAXN],n,m,p[MAXN],ret[MAXN],t[MAXN];
typedef long long ll;
ll k,d[MAXN];
inline void ksm(ll k)
{
	for(int i=1;i<=n;++i)ret[i]=i;
	while(k){
		if(k&1){
			for(int i=1;i<=n;++i)ret[i]=p[ret[i]];
		}
		for(int i=1;i<=n;++i)t[i]=p[i];
		for(int i=1;i<=n;++i)p[i]=t[p[i]];
		k>>=1;
	}
}
int v[MAXN],ans[MAXN];
int main()
{
	cin>>n;
	for(int i=1;i<=n;++i)scanf("%d",a+i),d[i]=a[i]-a[i-1],v[i]=i;
	cin>>m>>k;
	for(int i=1;i<=m;++i){
		int x;
		scanf("%d",&x);
		swap(v[x],v[x+1]);
	}
        for(int i=1;i<=n;++i)p[v[i]]=i;
	ksm(k);
        ll sum=0;
	for(int i=1;i<=n;++i)ans[ret[i]]=i;
	for(int i=1;i<=n;++i)
		printf("%lld\n",sum+=d[ans[i]]);
	return 0;
}
