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>
const int MN=100000+5;
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
template<typename T>T&IN(T&in){
	in=0;char c=getchar();int f=1;
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c))in=in*10+c-'0',c=getchar();
	return in*=f;
}
int n,m;ll k;
int fr[MN],st[MN],vis[MN];
ll a[MN],ans[MN];
void input(){
	IN(n);
	for(int i=1;i<=n;++i)IN(a[i]),fr[i]=i;
	for(int i=n;i>=1;--i)a[i]-=a[i-1];
	IN(m),IN(k);
	for(int i=1,x;i<=m;++i)IN(x),swap(fr[x],fr[x+1]);
	for(int i=1;i<=n;++i)if(!vis[i]){
		st[0]=0;
		for(int j=i;!vis[j];j=fr[j])vis[j]=1,st[++st[0]]=j;
		for(int j=1;j<=st[0];++j)ans[st[j]]=a[st[(j+k-1)%st[0]+1]];
	}
	for(int i=1;i<=n;++i)ans[i]+=ans[i-1],printf("%lld.0\n",ans[i]);
}
int main(){
	input();
	return 0;
}