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;
const ll MOD = 1e9 + 7;
const double PI = 3.141592653589793238;
const double EPS = 1e-10;
ll x[100000];
int perm[100000];
bool used[100000];
ll d[100000];
int main() {
	int N;
	cin >> N;
	for (int i = 0; i < N; i++) cin >> x[i];
	for (int i = 0; i + 1 < N; i++) perm[i] = i;
	int M;
	ll K;
	cin >> M >> K;
	for (int i = 0; i < M; i++) {
		int a;
		cin >> a;
		a--;
		swap(perm[a], perm[a - 1]);
	}
	for (int i = 0; i + 1 < N; i++) {
		if (used[i]) continue;
		vector<int> v;
		while (!used[i]) {
			v.push_back(i);
			used[i] = 1;
			i = perm[i];
		}
		for (int j = 0; j < v.size(); j++) {
			int k = v[(j + K) % v.size()];
			d[v[j]] = x[k + 1] - x[k];
		}
	}
	cout << x[0] << endl;
	ll now = x[0];
	for (int i = 0; i + 1 < N; i++) {
		now += d[i];
		cout << now << endl;
	}
}
