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 <cstdio>
using namespace std;

long long N, M, K, x[1000010], p[1000010], a, last, ans[100010], P[100010];

int main(){
	scanf("%lld", &N);
	for(int i=1; i<=N; i++)	scanf("%lld", &x[i]);
	for(int i=N; i; i--)	x[i]-=x[i-1], p[i]=i;
	scanf("%lld%lld", &M, &K);
	for(int i=1; i<=M; i++){
		scanf("%lld", &a);
		swap(p[a], p[a+1]);
	};
	for(int i=1; i<=N; i++)	ans[i]=i, P[i]=p[i];
	while(K){
		if(K&1)	
			for(int i=1; i<=N; i++)
				ans[i]=p[ans[i]];
		for(int i=1; i<=N; i++)
			P[i]=p[P[i]];
		for(int i=1; i<=N; i++)
			p[i]=P[i]; 
		K>>=1; 
	}
	for(int i=1; i<=N; i++){
		last+=x[ans[i]];
		printf("%lld\n", last);
	}
	return 0;
}