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 int long long
#define re register
#define inf 0x3f3f3f3f
#define mod 1000000007
using namespace std;
inline int read(){
	int x=0,w=0;char ch=getchar();
	while (!isdigit(ch))w|=ch=='-',ch=getchar();
	while (isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	return w?-x:x;
}
int a[100007],c[100007],v[100007],t[100007],n;
inline void ksm(int p){
	while (p){
		if (p&1){
			for (int i=1;i<=n;++i)t[i]=c[v[i]];
			for (int i=1;i<=n;++i)c[i]=t[i];
		}
		for (int i=1;i<=n;++i)t[i]=v[v[i]];
		for (int i=1;i<=n;++i)v[i]=t[i];
		p>>=1;
	}
}
signed main(){
    n=read();
    for(int i=1;i<=n;i++)a[i]=read(),c[i]=a[i]-a[i-1],v[i]=i;
    int m=read(),k=read();
    for(int i=1;i<=m;i++){
        int x=read();
        swap(v[x],v[x+1]);
    }
    ksm(k);
    double sum=0;
	for (int i=1;i<=n;++i){
		sum+=c[i];
		printf("%.1lf\n",sum);
	}
	return 0;
}