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
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;
#define N 200010
ll x[N],dis[N-1],child[N-1],last[N-1];
bool vis[N-1];
ll n,m,k;
int main(){
	cin>>n;
	lol(i,n)cin>>x[i];
	lol(i,n-1)dis[i]=x[i+1]-x[i];
	lol(i,n-1)child[i]=i,vis[i]=false;
	cin>>m>>k;
	lol(i,m){
		ll a;cin>>a;a--;
		swap(child[a-1],child[a]);
	}
	lol(i,n-1){
		if(vis[i]==true)continue;
		ll len=0,pnt=i;
		while(1){pnt=child[pnt],len++;if(pnt==i)break;}
		ll a=i,b=i;
		lol(u,k%len)b=child[b];
		while(1){last[a]=b,vis[a]=true;a=child[a],b=child[b];if(a==i)break;}
	}
	ll sum=x[0];
	cout<<sum<<endl;
	lol(i,n-1){
		sum+=dis[last[i]];
		cout<<sum<<endl;
	}
	return 0;
}
