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>

typedef long long ll;

const int kN = 1e5 + 5;

int n, m, a[kN], t[kN]; ll x[kN], d[kN], ans[kN], bas[kN];

void QPow(ll k) {
	while(k) {
		if(k & 1) {
			for(int i = 1; i <= n; ++i) t[i] = ans[bas[i]];
			for(int i = 1; i <= n; ++i) ans[i] = t[i];
		}
		for(int i = 1; i <= n; ++i) t[i] = bas[bas[i]];
		for(int i = 1; i <= n; ++i) bas[i] = t[i];
		k >>= 1;
	}
}

int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) scanf("%lld", x + i), ans[i] = i, bas[i] = i;
	ll k;
	scanf("%d%lld", &m, &k);
	for(int i = 1; i <= m; ++i) scanf("%d", a + i), std::swap(bas[a[i]], bas[a[i] + 1]);
	for(int i = 1; i <= n; ++i) d[i] = x[i] - x[i - 1];
	QPow(k);
	for(int i = 1; i <= n; ++i) t[i] = d[ans[i]];
	ll cur_ans = 0;
	for(int i = 1; i <= n; ++i) {
		cur_ans += t[i];
		printf("%lld.0\n", cur_ans);
	}
	return 0;
}