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 <bits/stdc++.h>
#define mod 1000000007
#define pb push_back
#define ll long long
using namespace std;
ll n,m,k;
vector<ll> arrange(vector<ll> prev, vector<ll> com){
	vector<ll> res(n+1);
	for(ll i=1;i<=n;i++){
		res[i] = prev[ com[i] ];
	}
	return res;
}
int main(){
	cin>>n;
	vector<ll> x(n+1),d(n+1),bas[63],st(n+1);
	bas[0].pb(0);
	for(ll i=1;i<=n;i++){
		cin>>x[i];
		bas[0].pb(i);
		st[i]=i;
	}
	cin>>m>>k;
	vector<ll> a(m+1);
	for(ll i=1;i<=m;i++){
		cin>>a[i];
	}
	d[0]=0;
	for(ll i=1;i<=n-1;i++){
		d[i]=x[i+1]-x[i];
	}
	for(ll i=1;i<=m;i++){
		swap(bas[0][a[i]-1],bas[0][a[i]]);
	}
	for(ll i=1;i<=60;i++){
		bas[i] = arrange(bas[i-1],bas[i-1]);
	}
	for(ll i=0;i<=60;i++){
		if(k%2==1){
			st = arrange(st,bas[i]);
		}
		k = k/2;
	}
	d = arrange(d, st);
	ll pos = x[1];
	for(ll i=0;i<n;i++){
		pos+=d[i];
		cout<<pos<<endl;
	}
	return 0;
}