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>
#define ll long long
using namespace std;
ll ans,n,m,k,a[200001],d[200001],b[200001],c[200001],nt[200001];
void ksm(){
	ll i;
	while(k){
		if(k&1){
			for(i=1;i<=n;i++)nt[i]=b[c[i]];
			for(i=1;i<=n;i++)b[i]=nt[i];
		}
		for(i=1;i<=n;i++)nt[i]=c[c[i]];
		for(i=1;i<=n;i++)c[i]=nt[i];
		k/=2;
	}
}
int main(){
	ll i,x;
	scanf("%lld",&n);
	for(i=1;i<=n;i++){
		scanf("%lld",&a[i]);
		d[i]=a[i]-a[i-1];
		b[i]=c[i]=i;
	}
	scanf("%lld%lld",&m,&k);
	for(i=1;i<=m;i++){
		scanf("%lld",&x);
		swap(c[x],c[x+1]);
	}
	ksm();
	for(i=1;i<=n;i++){
		ans+=d[b[i]];
		printf("%lld.0\n",ans);
	}
}
/*
思路：对于兔子x[i]，向左跳有2*x[i-1]-x[i]，向右跳有2*x[i+1]-x[i]，概率各位1/2，
得期望x[i-1]+x[i+1]-x[i]，令d[i] = x[i]-x[i-1]，
这条期望式子相当于swap(d[i], d[i+1])，那么连续执行K次，可参考快速幂的做法进行优化。O(NlogK)。
当然也可以O（n）

心得：没有想到用差分 
*/