Problem: 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

Solution (in CPP):

Solution:
	#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>

typedef long long LL;
const int MX=100011;

int p[MX],x[MX],a[MX];
int n,m;LL k;
LL ans[MX];

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&x[i]);
	for(int i=1;i<=n;i++)p[i]=i;
	scanf("%d%lld",&m,&k);
	for(int i=1;i<=m;i++)scanf("%d",&a[i]);for(int i=m;i;i--)std::swap(p[a[i]],p[a[i]+1]);
	static int t[MX];for(int i=1;i<=n;i++)t[i]=i;
	while(k){
		static int P[MX];
		if(k&1){for(int i=1;i<=n;i++)P[i]=p[t[i]];memcpy(t+1,P+1,sizeof(int)*n);}
		for(int i=1;i<=n;i++)P[i]=p[p[i]];memcpy(p+1,P+1,sizeof(int)*n);
		k>>=1;
	}
	for(int i=1;i<=n;i++)ans[t[i]]+=x[i]-x[i-1];
	for(int i=1;i<=n;i++)ans[i]+=ans[i-1];
	for(int i=1;i<=n;i++)printf("%lld\n",ans[i]);
	return 0;
}