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<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
typedef long long ll;
int gi() {
    int x=0,o=1;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int n,m,id[N],st[N],top=0;
ll a[N],ans[N],k;
bool vis[N];
int main() {
    cin>>n;
    for(int i=1;i<=n;i++) a[i]=gi(),id[i]=i;
    for(int i=n;i;i--) a[i]-=a[i-1];
    cin>>m>>k;
    for(int i=1,x;i<=m;i++) x=gi(),swap(id[x],id[x+1]);
    for(int i=1;i<=n;i++)
        if(!vis[i]) {
            top=0;
            for(int j=i;!vis[j];j=id[j]) st[++top]=j,vis[j]=1;
            for(int j=1;j<=top;j++) ans[st[j]]=a[st[(k+j-1)%top+1]];
        }
    for(int i=1;i<=n;i++) ans[i]+=ans[i-1],printf("%lld\n",ans[i]);
    return 0;
}
//orzgzy
//鸡贼明年进队超稳