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 MAX 100005
#define ll long long
#define P 998244353
using namespace std;

int n, m;
ll k, b[MAX];
double a[MAX], tmp[MAX];

void calc(ll k){
	while(k){
		if(k&1){
			for(int i = 1; i <= n; i++) tmp[i] = a[b[i]];
			for(int i = 1; i <= n; i++) a[i] = tmp[i];
		}
		for(int i = 1; i <= n; i++) tmp[i] = b[b[i]];
		for(int i = 1; i <= n; i++) b[i] = tmp[i];
		k >>= 1;
	}
}

int main()
{
	cin >> n;
	for(int i = 1; i <= n; i++){
		scanf("%lf", &a[i]);
		b[i] = i;
	}
	for(int i = n; i >= 1; i--){
		a[i] -= a[i-1];
	}
	cin >> m >> k;
	int x;
	for(int i = 1; i <= m; i++){
		scanf("%d", &x);
		swap(b[x], b[x+1]);
	}
	calc(k);
	double s = 0;
	for(int i = 1; i <= n; i++){
		s += a[i];
		printf("%.1lf\n", s);
	}
	
	return 0;
}