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<bits/stdc++.h>
using namespace std;

int n,id[100005],ans[100005],a[100005],tmp[100005],m;

long long k;

double output[100005];

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",a+i),id[i]=ans[i]=i;
	//d[i]记录每只兔子跳一次之后下一次跳的位置，ans[i]记录跳完k轮之后每只兔子的位置
	scanf("%d%lld",&m,&k);
	for(int i=1,x;i<=m;i++){
		scanf("%d",&x);
		swap(id[x],id[x+1]);
	}
	while(k){
		if(k&1){
			for(int i=1;i<=n;i++) tmp[i]=ans[id[i]];
			for(int i=1;i<=n;i++) ans[i]=tmp[i];
		}
		for(int i=1;i<=n;i++) tmp[i]=id[id[i]];
		for(int i=1;i<=n;i++) id[i]=tmp[i];
		k>>=1;
	}
	for(int i=1;i<=n;i++) output[i]=(double)(a[ans[i]]-a[ans[i]-1]);
	for(int i=1;i<=n;i++) printf("%.1f\n",output[i]+=output[i-1]);
}