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 <iomanip>
using namespace std;

const int N = 100 * 1000 + 5, LG = 61;

int dp[N][LG];
long double a[N];
long double ans[N];

long long get(long long x, long long y) {
	return (x >> y) & 1;
}


int main() {
	int n, m;
	long long k;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		dp[i][0] = i;
	}
	cin >> m >> k;
	for (int i = 0; i < m; i++) {
		int x;
		cin >> x;
		swap(dp[x][0], dp[x - 1][0]);
	}
	for (int j = 1; j < LG; j++) {
		for (int i = 1; i < n; i++) {
			dp[i][j] = dp[dp[i][j - 1]][j - 1];
		}
	}
	ans[1] = a[1];
	for (int i = 2; i <= n; i++) {
		int x = i - 1;
		for (int j = 0; j < LG; j++) {
			if (get(k, j)) {
				x = dp[x][j];
			}
		}
		ans[i] = (long long) ans[i - 1] + a[x + 1] - a[x];
	}
	cout << fixed << setprecision(7);
	for (int i = 1; i <= n; i++) {
		cout << ans[i] << endl;
	}
	return 0;
}