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>
#include <vector>

#define int long long
using namespace std;
const int N=100005;

int n,m;
int x[N],a[N],poi[N],cf[N],cnt;
long long k;
int block[N],num[N],vis[N];

vector<int>q[N];

void dfs(int x,int dep)
{
    num[x]=dep;vis[x]=1;block[x]=cnt;
    q[cnt].push_back(x);
    if(!vis[poi[x]]) dfs(poi[x],dep+1);
}
signed main()
{
    scanf("%lld",&n);
    for(int i=1;i<=n;++i) scanf("%lld",&x[i]);
    for(int i=1;i<n;++i) cf[i]=x[i+1]-x[i];
    scanf("%lld%lld",&m,&k);
    for(int i=1;i<n;++i)poi[i]=i;
    for(int i=1;i<=m;++i) {scanf("%lld",&a[i]);swap(poi[a[i]],poi[a[i]-1]);}
    for(int i=1;i<n;++i) if(!vis[i]) cnt++,dfs(i,0);
    
    printf("%lld\n",x[1]);
    int now=x[1];
    for(int i=1;i<n;i++)
    {
        now+=cf[q[block[i]][(num[i]+k)%q[block[i]].size()]];
        printf("%lld\n",now);
    }
}