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 <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10;

int n, m, ans[N], ord[N];
long long qn, a[N], t[N];

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld", a + i);
		ord[i] = ans[i] = i;
	}
	for (int i = n; i >= 1; --i) {
		a[i] -= a[i - 1];
	}
	scanf("%d%lld", &m, &qn);
	for (int i = 0; i < m; ++i) {
		int x;
		scanf("%d", &x);
		swap(ord[x], ord[x + 1]);
	}
	for (; qn; qn >>= 1) {
		if (qn & 1) {
			for (int i = 1; i <= n; ++i) {
				ans[i] = ord[ans[i]];
			}
		}
		for (int i = 1; i <= n; ++i) {
			t[i] = ord[ord[i]];
		}
		for (int i = 1; i <= n; ++i) {
			ord[i] = t[i];
		}
	}
	for (int i = 1; i <= n; ++i) {
		t[i] = a[ans[i]];
	}
	for (int i = 1; i <= n; ++i) {
		a[i] = t[i] + a[i - 1];
		printf("%lld.0\n", a[i]);
	}
}