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;

typedef long long ll;
const int maxn=1e5+10;
int n,m,a[maxn],x[maxn],d[maxn],p[maxn],fa[63][maxn];
ll k,ans[maxn];

int main(){
	cin>>n;
	for(int i=1;i<=n;++i)
		cin>>a[i];
	cin>>m>>k;
	for(int i=1;i<=m;++i)
		cin>>x[i];
	for(int i=1;i<n;++i)
		d[i]=a[i+1]-a[i];
	for(int i=1;i<n;++i)
		p[i]=i;
	for(int i=1;i<=m;++i)
		swap(p[x[i]-1],p[x[i]]);
	for(int i=1;i<n;++i)
		fa[0][i]=p[i];
	for(int i=1;i<63;++i)
		for(int j=1;j<n;++j)
			fa[i][j]=fa[i-1][fa[i-1][j]];
	ans[1]=a[1];
	for(int i=1;i<n;++i){
		int pos=i;
		for(int j=62;~j;--j)
			if(k>>j&1ll)
				pos=fa[j][pos];
		ans[i+1]=ans[i]+d[pos];
	}
	for(int i=1;i<=n;++i)
		printf("%lld\n",ans[i]);
	return 0;
}