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>
using namespace std;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000000000


int main(){
	int N;
	cin>>N;
	vector<long long> x(N);
	vector<long long> d(N-1);
	for(int i=0;i<N;i++){
		cin>>x[i];
		if(i!=0){
			d[i-1] = x[i] - x[i-1];
		}
	}
	
	
	
	long long M,K;
	cin>>M>>K;
		
	vector<long long> v(N-1);
	for(int i=0;i<N-1;i++){
		v[i]=i;
	}
	
	for(int i=0;i<M;i++){
		int b;
		cin>>b;
		swap(v[b-2],v[b-1]);
	}
	
	
	while(K!=0){
		if(K%2==1){
			
			vector<long long> dd(N-1);
			for(int i=0;i<N-1;i++){
				dd[i] = d[v[i]];
			}
			d=dd;
		}
		vector<long long> c=v;
		for(int i=0;i<N-1;i++){
			c[i] = v[c[i]];
		}
		v=c;
		K/=2;
	}
	
	for(int i=1;i<N;i++){
		x[i] = x[i-1] + d[i-1];
	}
	
	for(int i=0;i<N;i++){
		cout<<x[i]<<endl;
	}
	
	
	
	
	
	return 0;
}