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<vector>
#include<cstring>
#include<algorithm>
#define int long long
using namespace std;
const int N=100005;
int X[N],cha[N],p[N],cnt[N];
int a[N],dfn[N],c[N];
vector<long long> O[N];
int n,m,k,num,s;
void find(int x,int len){
    dfn[x]=len,c[x]=num;
    cnt[num]++;
    O[num].push_back(x);
    if(!c[p[x]]) find(p[x],len+1);
}
signed main (){
    scanf("%lld",&n);
    for(int i=1;i<=n;i++)
    scanf("%lld",&X[i]),cha[i-1]=X[i]-X[i-1];
    for(int i=1;i<=n;i++) 
    p[i]=i;
    
	scanf("%lld%lld",&m,&k);
    for(int i=1;i<=m;i++)
    scanf("%lld",&a[i]),
	swap(p[a[i]],p[a[i]-1]);
	
    for(int i=1;i<=n;i++)if(!dfn[i]) 
	num++,find(i,0);
    s=X[1],printf("%.1lf\n",1.0*s);
    for(int i=1;i<n;i++){
    	int pos=(dfn[i]+k)%O[c[i]].size();
		printf("%.1lf\n",1.0*(s+=cha[O[c[i]][pos]]));
	}
}