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 <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
	int N, M; long long K;
	cin >> N;
	vector<int> X(N);
	for (int i = 0; i < N; ++i) cin >> X[i];
	cin >> M >> K;
	vector<int> perm(N - 1);
	for (int i = 0; i < N - 1; ++i) perm[i] = i;
	for (int i = 0; i < M; ++i) {
		int A;
		cin >> A;
		swap(perm[A - 2], perm[A - 1]);
	}
	vector<int> p(N - 1);
	for (int i = 0; i < N - 1; ++i) p[i] = i;
	for (int i = 0; i < 60; ++i) {
		if ((K >> i) & 1) {
			vector<int> nxtp(N - 1);
			for (int j = 0; j < N - 1; ++j) {
				nxtp[j] = p[perm[j]];
			}
			p = nxtp;
		}
		vector<int> nxtperm(N - 1);
		for (int j = 0; j < N - 1; ++j) {
			nxtperm[j] = perm[perm[j]];
		}
		perm = nxtperm;
	}
	long long ans = X[0];
	for (int i = 0; i < N; ++i) {
		cout << ans << '\n';
		if (i != N - 1) ans += X[p[i] + 1] - X[p[i]];
	}
	return 0;
}