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;
typedef long long ll;
int A[100005],id[100005],vis[100005];
int cur[100005],id2[100005];ll k;
void sv(int a)
{
	int t = id[a];
	int c = 0;
	cur[++ c] = a;vis[a] = 1; 
	while(t != a)
		cur[++ c] = t,vis[t] = 1,t = id[t];
	int r = k%c;
	for(int i = 1;i <= c;i ++)
		id2[cur[i]] = cur[(i+r-1)%c+1];
}
int main()
{
	int n,m;
	scanf("%d",&n);
	for(int i = 1;i <= n;i ++)
		scanf("%d",&A[i]),id[i] = i;
	scanf("%d%lld",&m,&k);
	for(int i = 1;i <= m;i ++)
	{
		int a;
		scanf("%d",&a);
		swap(id[a],id[a+1]);
	}
	for(int i = 1;i <= n;i ++)
		if(!vis[i])
			sv(i);
	ll t = A[1];printf("%lld.0\n",t); 
	for(int i = 2;i <= n;i ++)
	{
		t += A[id2[i]]-A[id2[i]-1];
		printf("%lld.0\n",t);
	}
	return 0;
}