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<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
typedef long long ll;
const int maxn=100000+10;
ll x[maxn],d[maxn],c[maxn];
int ans[maxn],now[maxn],o[maxn],sta[maxn];
int i,j,l,t,n,m,top;
ll k;
void mult(int *a,int *b,int *c){
    int i;
    fo(i,1,n) o[i]=a[b[i]];
    fo(i,1,n) c[i]=o[i];
}
int main(){
    scanf("%d",&n);
    fo(i,1,n){
        scanf("%lld",&x[i]);
        d[i]=x[i]-x[i-1];
        ans[i]=now[i]=i;
    }
    scanf("%d%lld",&m,&k);
    fo(i,1,m){
        scanf("%d",&t);
        swap(now[t],now[t+1]);
    }
    top=0;
    while (k){
        sta[++top]=k%2;
        k/=2;
    }
    while (top){
        mult(ans,ans,ans);
        if (sta[top--]) mult(ans,now,ans);
    }
    fo(i,1,n) c[i]=d[ans[i]];
    fo(i,1,n) x[i]=x[i-1]+c[i];
    fo(i,1,n) printf("%lld.0\n",x[i]);
}