Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=100005;
typedef long long ll;
int n,m,p[N];
ll k;
ll a[N];
ll dist[N];
bool vis[N];
ll ans[N];
int st[N],top;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
        p[i]=i;
        dist[i]=a[i]-a[i-1];
    }
    scanf("%d%lld",&m,&k);
    for(int i=1;i<=m;i++){
        int x;
        scanf("%d",&x);
        swap(p[x],p[x+1]);
    }
    for(int i=1;i<=n;i++)
        if(!vis[i]){
            top=0;
            for(int j=i;!vis[j];j=p[j])
                st[++top]=j,vis[j]=1;
            for(int j=1;j<=top;j++)
                ans[st[j]]=dist[st[(k+j-1)%top+1]];
        }
    for(int i=1;i<=n;i++)
        ans[i]+=ans[i-1];
    for(int i=1;i<=n;i++)
        printf("%lld\n",ans[i]);
}