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 MAXN = 100005;
ll X[MAXN],Y[MAXN];
int g[MAXN],fn[MAXN];
bool vis[MAXN];
int main()
{
	ios::sync_with_stdio(0);
	int n;
	cin >> n;
	for(int i=1;i<=n;i++)
		cin >> X[i];
	for(int i=n;i;i--) {
		X[i] = X[i] - X[i-1];
		g[i] = i;
	}
	int m;
	ll K;
	cin >> m >> K;
	for(int i=1;i<=m;i++) {
		int pos;
		cin >> pos;
		swap(g[pos] , g[pos+1]);
	}
	
	for(int i=1;i<=n;i++)
		if(vis[i] == false) {
			static int st[MAXN];
			int cnt = 0;
			for(int j=i;vis[j] == false;j=g[j]) {
				st[cnt++] = j;
				vis[j] = true;
			}
			for(int j=0;j<cnt;j++)
				fn[st[j]] = st[(j+K)%cnt];
		}
	
	for(int i=1;i<=n;i++)
		Y[i] = X[fn[i]];
	for(int i=1;i<=n;i++) {
		Y[i] += Y[i-1];
		cout << Y[i] << ".0\n";
	}
}