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<algorithm>
#include<cmath>
using namespace std;

typedef long long ll;
const int MAXN=1e5+5;

ll n,m,k;
ll a[MAXN],ans[MAXN],id[MAXN];
ll sta[MAXN],top;
bool vis[MAXN];

int main(){

	scanf("%lld",&n);
	for(int i=1;i<=n;i++) scanf("%lld",&a[i]),id[i]=i;
	for(int i=n;i>=1;i--) a[i]-=a[i-1];
	scanf("%lld%lld",&m,&k);
	for(int i=1;i<=m;i++){
	    int x;
        scanf("%d",&x);
        swap(id[x],id[x+1]);
	}
	for(int i=1;i<=n;i++){
	    if(vis[i]) continue;
	    top=0;
	    for(int j=i;j<=n&&!vis[j];j=id[j]) sta[++top]=j,vis[j]=1;
	    for(int j=1;j<=top;j++) ans[sta[j]]=a[sta[(j+k-1)%top+1]];
	}
	for(int i=1;i<=n;i++)
		ans[i]+=ans[i-1],printf("%lld\n",ans[i]);

	fclose(stdin);
	fclose(stdout);
}