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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll;
#define rep(i, a, b) for (int i = a; i <= b; ++ i)
const int maxn = 100010;
using namespace std;
int n, a[maxn], f[maxn], w[maxn], m, x, tmp[maxn]; ll k;
int main() {
	scanf("%d", &n);
	rep(i, 1, n) scanf("%d", &a[i]);
	rep(i, 2, n) f[i] = i;
	scanf("%d%lld", &m, &k);
	rep(i, 1, m) scanf("%d", &x), swap(f[x], f[x + 1]);
	rep(i, 2, n) w[i] = f[i], f[i] = i;
	for (; k; k /= 2) {
		if (k & 1LL) rep(i, 2, n) f[i] = w[f[i]];
		rep(i, 2, n) tmp[i] = w[i];
		rep(i, 2, n) w[i] = tmp[w[i]];
	} 
	printf("%d.0\n", a[1]); ll las = a[1];
	for (int i = 2; i <= n; ++ i) 
		las += a[f[i]] - a[f[i] - 1], printf("%lld.0\n", las);
	return 0;
}