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>
int N, M, ans[100001], bit[100001], tmp[100001];
long long K, a[100001], s[100001];
void MUL(int *a, int *b, int *c)
{
	for (int i = 0; i <= N; i++)
		tmp[i] = a[b[i]];
	std::copy(tmp, tmp + N + 1, c);
}
int main()
{
	scanf("%d", &N);
	for (int i = 1; i <= N; i++)
		scanf("%lld", a + i);
	for (int i = 0; i <= N; i++)
	{
		bit[i] = i;
		ans[i] = i;
	}
	scanf("%d%lld", &M, &K);
	for (int i = 1, u; i <= M; i++)
	{
		scanf("%d", &u);
		std::swap(bit[u - 1], bit[u]);
	}
	for (; K; K >>= 1)
	{
		if (K & 1)
			MUL(ans, bit, ans);
		MUL(bit, bit, bit);
	}
	for (int i = 1; i < N; i++)
		a[i] -= a[i + 1];
	for (int i = 1; i <= N; i++)
		s[i] = a[ans[i]];
	for (int i = N - 1; i; i--)
		s[i] += s[i + 1];
	for (int i = 1; i <= N; i++)
		printf("%lld.0\n", s[i]);
	return 0;
}