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>
using namespace std; 
int n,c,x[100105],m,d[100105],ans[100105],tmp[100105];
long long k;
double a[100105];
int main(){
    scanf("%d",&n);
    for(int i = 1;i <= n;i++){
    	scanf("%d",&x[i]);
	}
    scanf("%d%lld",&m,&k);
    for(int i = 1;i <= n;i++){
    	d[i] = i;
		ans[i] = i;
	}
    for(int i = 1;i <= m;i++){
        scanf("%d",&c);
        swap(d[c],d[c+1]);
    }
    while(k){
        if(k&1){
            for(int i = 1;i <= n;i++) tmp[i] = ans[d[i]];
            for(int i = 1;i <= n;i++) ans[i] = tmp[i];
        }
        for(int i = 1;i <= n;i++) tmp[i] = d[d[i]];
        for(int i = 1;i <= n;i++) d[i] = tmp[i];
        k >>= 1;
    }
    for(int i = 1;i <= n;i++) a[i] = x[ans[i]]-x[ans[i]-1];
    for(int i = 1;i <= n;i++){
    	printf("%0.1lf\n",a[i] += a[i-1]);
	}
    return 0;
}