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;

typedef long long ll;
const int N = 1e5 + 100, LOG = 62;

ll n, m, k, dex, x[N], d[N], is[N], go[N][LOG];

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> x[i];
		if(i)
			d[i] = x[i] - x[i - 1], go[i][0] = i;
	}
	cin >> m >> k;
	while(m--) {
		cin >> dex;
		dex--;
		swap(go[dex][0], go[dex + 1][0]);
	}
	for (int j = 1; j < LOG; j++)
		for (int i = 0; i < n; i++)
			go[i][j] = go[go[i][j - 1]][j - 1];
	iota(is, is + n, 0);
	for (int j = LOG - 1; ~j; j--)
		if(k >= (1LL << j)) {
			k -= 1LL << j;
			for (int i = 1; i < n; i++)
				is[i] = go[is[i]][j];
		}
	ll now = x[0];
	cout << now << ' ';
	for (int i = 1; i < n; i++) {
		now += d[is[i]];
		cout << now << ' ';
	}
	return 0;
}