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
#pragma GCC optimize("trapv")
#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
const int maxn = 1<<17;
ll n, m, k;
vector<ll> a, d, p;
void apply(vector<ll> a, vector<ll> &b) {
	vector<ll> res(a.size());
	for(int i = 0; i < a.size(); i++) res[i] = b[a[i]];
	b = res;
}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n;
	a.resize(n);
	for(auto &i : a) cin >> i;
	for(int i = 1; i < n; i++) d.push_back(a[i]-a[i-1]);
	cin >> m >> k;
	p.resize(n-1);
	iota(all(p), 0);
	auto r = p;
	for(int t, i = 0; i < m; i++) {
		cin >> t;
		swap(p[t-1], p[t-2]);
	}
	while(k) {
		if(k&1) apply(p, r);
		apply(p, p), k>>=1;
	}
	apply(r, d);
	cout << a[0] << '\n';
	for(auto i : d) cout << (a[0] += i) << "\n";
}
