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;
typedef long long ll;
     
const int maxn(1e5 + 5);
     
int n, m, b[maxn], nxt[maxn], que[maxn], len, vis[maxn];
ll k, f[maxn], g[maxn];
     
int main() {
	int i, j;
	scanf("%d", &n);
	for (i = 1; i <= n; ++i) scanf("%lld", &f[i]), b[i] = i;
	for (i = n; i; --i) f[i] -= f[i - 1];
	scanf("%d%lld", &m, &k);
	for (i = 1; i <= m; ++i) scanf("%d", &j), swap(b[j], b[j + 1]);
	for (i = 1; i <= n; ++i)
		if (!vis[i]) {
			vis[i] = 1, j = b[i], que[len = 1] = i;
			while (!vis[j]) que[++len] = j, vis[j] = 1, j = b[j];
			for (j = 1; j <= len; ++j) nxt[que[j]] = que[(k + j - 1) % len + 1];
		}
	for (i = 1; i <= n; ++i) g[i] = f[nxt[i]];
	for (i = 1; i <= n; ++i) g[i] += g[i - 1], printf("%lld\n", g[i]);
	return 0;
}