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>
#define maxn 100010
#define ll long long
using namespace std;

int n,m,vis[maxn];

ll ans[maxn],a[maxn],k,cnt,q[maxn],pos[maxn];

template <class T>
void read(T &x){
	char s=getchar();int f=0;x=0;
	while(s<'0'||s>'9'){if(s=='-') f=1;s=getchar();}
	while(s>='0'&&s<='9') x=(x<<3)+(x<<1)+s-'0',s=getchar();
	x=f?-x:x;
}

int main(){
	read(n);
	for(int i=1;i<=n;i++){
		read(a[i]);
		pos[i]=1ll*i;
	}
	read(m);read(k);
	for(int i=n;i>=1;i--) a[i]=a[i]-a[i-1];
	for(int i=1,x;i<=m;i++){
		read(x);
		swap(pos[x],pos[x+1]);
	}
	for(int i=1;i<=n;i++){
		ll p=i;cnt=0;
		if(vis[i]) continue;
		q[++cnt]=p;
		vis[p]=1;
		p=pos[p];
		while(p!=i){
			q[++cnt]=p;
			vis[p]=1;
			p=pos[p];
		}
		for(int j=1;j<=cnt;j++) ans[q[j]]=a[q[(j+k-1)%cnt+1]];
	}
	for(int i=1;i<=n;i++){
		ans[i]+=ans[i-1];
		printf("%lld.0\n",ans[i]);
	}
	return 0;
}