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 <iostream>
#include <algorithm>
using namespace std;
const int N=100005;
#define ll long long
int pos[N],a[N];
ll p[N],d[N];
int to[N][64];
int main (){
	int n;scanf ("%d",&n);
	for (int i=1;i<=n;i++) scanf ("%lld",&p[i]),d[i]=p[i]-p[i-1];
	int m;ll k;
	scanf ("%d%lld",&m,&k);
	for (int i=1;i<=n;i++) a[i]=i;
	for (int i=1,x;i<=m;i++) {scanf ("%d",&x);swap(a[x],a[x+1]);}
	for (int i=1;i<=n;i++) to[a[i]][0]=i;
	for (int i=1;i<=60;i++)
		for (int j=1;j<=n;j++)
			to[j][i]=to[to[j][i-1]][i-1];
	for (int i=1;i<=n;i++){
		int now=i;
		for (int j=0;j<=60;j++)
			if (k&(1ll<<j)) now=to[now][j];
		pos[now]=i;
	}
	for (int i=1;i<=n;i++) p[i]=p[i-1]+d[pos[i]],printf ("%lld\n",p[i]);
	return 0;
}