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>

typedef long long LL;
const int MN = 100005;

int N, M; LL K;
LL x[MN];
int per[MN], vis[MN], stk[MN], tp;

int main() {
	scanf("%d", &N);
	for (int i = 1; i <= N; ++i) scanf("%lld", &x[i]);
	for (int i = N; i >= 2; --i) x[i] -= x[i - 1];
	for (int i = 1; i <= N; ++i) per[i] = i;
	scanf("%d%lld", &M, &K);
	while (M--) {
		int p;
		scanf("%d", &p);
		std::swap(per[p], per[p + 1]);
	}
	for (int i = 1; i <= N; ++i) if (!vis[i]) {
		tp = 0;
		int u = i;
		do vis[u] = 1, stk[tp++] = u, u = per[u]; while (u != i);
		for (int j = 0; j < tp; ++j) per[stk[j]] = stk[(j + K) % tp];
	}
	LL s = 0;
	for (int i = 1; i <= N; ++i) s += x[per[i]], printf("%lld\n", s);
	return 0;
}