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
#include<bits/stdc++.h>
using namespace std;
#define MAXN	100005
bool visited[MAXN];
int n, m, num[MAXN], tmp[MAXN], dest[MAXN];
long long k, x[MAXN], delta[MAXN], ans[MAXN];
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &x[i]);
		delta[i] = x[i] - x[i - 1];
		num[i] = i;
	}
	scanf("%d%lld", &m, &k);
	for (int i = 1; i <= m; i++) {
		int x;
		scanf("%d", &x);
		swap(num[x], num[x + 1]);
	}
	for (int i = 1; i <= n; i++) {
		if (visited[i]) continue;
		int tot = 0, pos = i;
		while (!visited[pos]) {
			visited[pos] = true;
			tmp[tot++] = pos;
			pos = num[pos];
		}
		for (int j = 0; j < tot; j++)
			dest[tmp[j]] = tmp[(j + k) % tot];
	}
	for (int i = 1; i <= n; i++)
		ans[i] = delta[dest[i]];
	for (int i = 1; i <= n; i++) {
		ans[i] += ans[i - 1];
		printf("%lld.0\n", ans[i]);
	}
	return 0;
}