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<cstdlib>
#include<cstring>
#include<algorithm>
#define fo(i,j,k) for(int i=j;i<=k;++i)
#define fd(i,j,k) for(int i=j;i>=k;--i)
#define ll long long
using namespace std;
const int N=1e5+10;
int p[N],q[N];
ll a[N],an[N];
bool vis[N];
int z[N];
int main()
{
	int n;
	scanf("%d",&n);
	fo(i,1,n) scanf("%lld",&a[i]);
	fd(i,n,1) a[i]-=a[i-1],p[i]=i;
	int m;
	ll k;
	scanf("%d %lld",&m,&k);
	while(m--){
		int x;
		scanf("%d",&x);
		swap(p[x],p[x+1]);
	}
	fo(i,1,n) if(!vis[i]){
		z[0]=0;
		for(int j=i;!vis[j];j=p[j]) z[++z[0]]=j,vis[j]=1;
		int t=1-k%z[0];
		t<=0?t+=z[0]:t;
		fo(i,1,z[0]) q[z[i]]=z[t],t=t%z[0]+1;
	}
	fo(i,1,n) an[q[i]]=a[i];
	fo(i,1,n) an[i]+=an[i-1],printf("%lld.0\n",an[i]);
}