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 <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
using namespace std;
const int maxn=1e5+10;
typedef long long ll;
int n,f[maxn][60],dsy[maxn],fxk[maxn],m,x;ll ans,a[maxn],p[maxn],k; 
int main(){
	scanf("%d",&n);
	rep(i,1,n) scanf("%lld",&a[i]);
	rep(i,1,n-1) p[i]=a[i+1]-a[i],f[i][0]=i,fxk[i]=i;
	scanf("%d%lld",&m,&k);
	rep(i,1,m) scanf("%d",&x),swap(f[x-1][0],f[x][0]);
	rep(i,1,59) rep(j,1,n-1) f[j][i]=f[f[j][i-1]][i-1];
	per(i,59,0) if(k&(1ll<<i)){
		rep(j,1,n-1) dsy[j]=f[fxk[j]][i];
		rep(j,1,n-1) fxk[j]=dsy[j];
	}
	printf("%lld.0\n",a[1]);ans=a[1];
	rep(i,1,n-1) ans+=p[fxk[i]],printf("%lld.0\n",ans);
	return 0;
}