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 <iostream>
#include <algorithm>

using namespace std;

int x[100001], d[100001];
int a[100001], pos[70][100001];

int main() {
	int n, m;
	long long k;

	cin >> n;
	for (int i = 1; i <= n; ++i) cin >> x[i];

	cin >> m >> k;
	for (int i = 1; i <= m; ++i) cin >> a[i];

	for (int i = 1; i < n; ++i) d[i] = i;

	for (int i = 1; i <= m; ++i)
		swap(d[a[i] - 1], d[a[i]]);
	for (int i = 1; i < n; ++i)
		pos[0][d[i]] = i;
	
	for (int i = 1; i <= 61; ++i)
		for (int j = 1; j < n; ++j)
			pos[i][j] = pos[i - 1][pos[i - 1][j]];

	for (int i = 1; i < n; ++i) {
		int p = i;
		long long t = k;
		for (int j = 61; j >= 0; --j)
			if (t >= (1LL << j)) {
				t -= (1LL << j);
				p = pos[j][p];
			}
		d[p] = x[i + 1] - x[i];
	}

	long long sum = x[1];
	cout << sum << '\n';
	for (int i = 1; i < n; ++i) {
		sum += d[i];
		cout << sum << '\n';
	}

	return 0;
}