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 N 100005
using namespace std;
int n,m;
long long k;
long long x[N];
int v[N],p[N],vis[N],fin[N];
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%lld",&x[i]);
    for(int i=1;i<n;i++) v[i]=x[i+1]-x[i],p[i]=i;
    scanf("%d%lld",&m,&k);
    for(int i=1,t;i<=m;i++){
        scanf("%d",&t);
        swap(p[t-1],p[t]);
    }
    for(int i=1;i<n;i++)
        if(vis[i]==0){
            vector<int> vec;vec.push_back(i);
            for(int j=p[i];j!=i;j=p[j]) vec.push_back(j);
            int num=k%vec.size();
            for(int j=0;j<vec.size();j++){
                fin[vec[j]]=vec[(j+num)%vec.size()];
                vis[vec[j]]=1;
            }
        }
    for(int i=1;i<n;i++) fin[i]=v[fin[i]];
    for(int i=2;i<=n;i++) x[i]=x[i-1]+fin[i-1];
    for(int i=1;i<=n;i++) printf("%lld\n",x[i]);
    return 0;
}