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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
 
#define For(i, j, k) for(int i = j; i <= k; i++)
 
const int N = 100010;
 
double A[N];
int n, m, pos[N], op[N], C[N], P[N], tmp[N];
long long k;
 
void mul(int *u, int *v){
	For(i, 1, n - 1) tmp[i] = v[u[i]];
	For(i, 1, n - 1) u[i] = tmp[i];
}
 
int main(){
	scanf("%d", &n);
	For(i, 1, n) scanf("%lf", &A[i]), pos[i] = C[i] = i;
	scanf("%d%lld", &m, &k);
	For(i, 1, m) scanf("%d", &op[i]), std::swap(pos[op[i]], pos[op[i] - 1]);
	For(i, 1, n - 1) P[pos[i]] = i;
 
	while(k){
		if(k & 1) mul(C, P);
		mul(P, P);
		k >>= 1;
	}
 
	For(i, 1, n - 1) pos[C[i]] = i;
 
	double now = A[1];
	printf("%.0lf\n", now);
	For(i, 1, n - 1) printf("%.0lf\n", now += A[pos[i] + 1] - A[pos[i]]);
 
	return 0;
}