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
// Rabbit Exercise
// * frank_c1
// * 2017 / 11/ 07

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

const int maxn = (int)(1e5) + 5;
int p[maxn], pt[maxn], tmp[maxn];
LL c[maxn], nc[maxn];

int main() {
	int n; scanf("%d", &n);
	for (int i = 1; i <= n; ++i) scanf("%lld", &c[i]);
	for (int i = n; i >= 1; --i) c[i] -= c[i - 1], p[i] = i;
	int m; LL K; scanf("%d%lld", &m, &K);
	for (int i = 1; i <= m; ++i) {
		int x; scanf("%d", &x); swap(p[x], p[x + 1]);
	}
	for (int i = 1; i <= n; ++i) pt[i] = i;
	for (; K; K >>= 1) {
		if (K & 1) {
			memcpy(tmp, pt, sizeof(pt));
			for (int i = 1; i <= n; ++i) pt[i] = tmp[p[i]];
		}
		memcpy(tmp, p, sizeof(p));
		for (int i = 1; i <= n; ++i) p[i] = tmp[tmp[i]];
	} LL ret = 0;
	for (int i = 1; i <= n; ++i) nc[i] = c[pt[i]];
	for (int i = 1; i <= n; ++i) printf("%lld\n", (nc[i] += nc[i - 1]));
	return 0;
}