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<stdio.h>
#include<algorithm>
using namespace std;
int x[110000];
int p[110000];
int t[110000];
int s[110000];
int tmp[110000];
int r[110000];
long long sum[110000];
int main(){
	int a;scanf("%d",&a);
	for(int i=0;i<a;i++)scanf("%d",x+i);
	int b;
	long long c;
	scanf("%d%lld",&b,&c);
	for(int i=0;i<b;i++){
		scanf("%d",p+i);p[i]--;
	}
	for(int i=0;i<=a;i++)t[i]=i;
	for(int i=0;i<b;i++){
		swap(t[p[i]],t[p[i]+1]);
	}
	for(int i=0;i<=a;i++)s[t[i]]=i;
	for(int i=0;i<=a;i++)r[i]=i;
	while(c){
		if(c%2){
			for(int i=0;i<=a;i++)tmp[i]=s[r[i]];
			for(int i=0;i<=a;i++)r[i]=tmp[i];
		}
		c/=2;
		for(int i=0;i<=a;i++)tmp[i]=s[s[i]];
		for(int i=0;i<=a;i++)s[i]=tmp[i];
	}
	for(int i=0;i<a;i++){
		sum[r[i]]+=x[i];
		sum[r[i+1]]-=x[i];
	}
	for(int i=0;i<a;i++){
		if(i)sum[i]+=sum[i-1];
		printf("%lld\n",sum[i]);
	}
}