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 int long long
using namespace std;
int n,m,x[100050],a[100050],p[100050],s[100050],cnt,k;
int be[100050],num[100050];
bool vis[100050]={0};
vector<int>q[100050];
void dfs(int x,int dep){
	num[x]=dep,vis[x]=1,be[x]=cnt;
	q[cnt].push_back(x);
	if(!vis[p[x]]) dfs(p[x],dep+1);
}
signed main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		scanf("%lld",&x[i]);
	}
	for(int i=1;i<n;i++){
		s[i]=x[i+1]-x[i];
	}
	cin>>m>>k;
	for(int i=1;i<n;i++) p[i]=i;
	for(int i=1;i<=m;i++){
		scanf("%lld",&a[i]);
		swap(p[a[i]],p[a[i]-1]);
	}
	for(int i=1;i<n;i++){
		if(!vis[i]){
			cnt++;
			dfs(i,0);
		}
	}
	cout<<x[1]<<endl;
	int now=x[1];
	for(int i=1;i<n;i++){
		now+=s[q[be[i]][(num[i]+k)%q[be[i]].size()]];
		cout<<now<<endl;
	}
	return 0;
}