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;
typedef unsigned long long ul;
#define sz(x) ((int)(x).size())
#define rep(i,l,r) for(int i=(l),I=(r);i<I;++i)
//-------head-------
const int N = 1e5 + 7;
int n, m, a[N], x[N], p[N], t[N];
ll k, ans[N];
int main() {
	scanf("%d", &n);
	rep(i, 1, n + 1) scanf("%d", &x[i]);
	scanf("%d%lld", &m, &k);
	rep(i, 1, n + 1) a[i] = i;
	rep(i, 0, m) {
		int w;
		scanf("%d", &w);
		swap(a[w], a[w + 1]);
	}
	rep(i, 1, n + 1) p[a[i]] = i, a[i] = i;
	rep(z, 0, 60) {
		if (k >> z & 1) {
			rep(i, 1, n + 1) t[p[i]] = a[i];
			rep(i, 1, n + 1) a[i] = t[i];
		}
		rep(i, 1, n + 1) t[i] = p[p[i]];
		rep(i, 1, n + 1) p[i] = t[i];
	}
	ans[1] = x[1];
	rep(i, 2, n + 1) ans[i] = ans[i - 1] + x[a[i]] - x[a[i] - 1];
	rep(i, 1, n + 1) printf("%lld\n", ans[i]);
	return 0;
}