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<algorithm>
using namespace std;
const int MAXN = 100000;
typedef long long ll;
int N, M; ll K;
int x[MAXN + 5];
int a[MAXN + 5], b[MAXN + 5], res[MAXN + 5];
void Pow(ll p) {
	for(int i=1;i<N;i++)
		res[i] = i;
	while( p ) {
		if( p & 1 ) {
			for(int i=1;i<N;i++)
				res[i] = a[res[i]];
		}
		p >>= 1;
		for(int i=1;i<N;i++)
			b[i] = a[a[i]];
		for(int i=1;i<N;i++)
			a[i] = b[i];
	}
}
int main() {
	scanf("%d", &N);
	for(int i=0;i<N;i++)
		scanf("%d", &x[i]);
	for(int i=N-1;i>0;i--)
		x[i] = x[i] - x[i-1];
	scanf("%d%lld", &M, &K);
	for(int i=1;i<N;i++)
		a[i] = i;
	for(int i=1;i<=M;i++) {
		int k;
		scanf("%d", &k);
		swap(a[k-1], a[k]);
	}
	Pow(K);
	ll nw = x[0];
	printf("%lld\n", nw);
	for(int i=1;i<N;i++) {
		nw = nw + x[res[i]];
		printf("%lld\n", nw);
	}
}