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
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

bool vis[100050];
ll w[100050], nxt[100050], que[100050], output[100050];

int main(){
	int n, m;
	scanf("%d", &n);
	for(int i = 1; i <= n; i++) scanf("%lld", &w[i]), nxt[i] = i;
	for(int i = n; i >= 1; i--) w[i] -= w[i-1];
	ll k;
	ll a;
	scanf("%d %lld", &m, &k);
	for(int i = 1; i <= m; i++) scanf("%lld", &a), swap(nxt[a], nxt[a+1]);
	for(int i = 1; i <= n; i++){
		if(!vis[i]) {
			int now = i, cnt = 0;
			while(!vis[now]){
				vis[now] = 1;
				que[++cnt] = now;
				now = nxt[now];
			}
			for(int j = 1; j <= cnt; j++){
				ll t = (0ll+j+k-2)%cnt+1;
				output[que[j]] = w[nxt[que[t]]];
			}
		}
	}
	ll p = 0;
	for(int i = 1; i <= n; i++) p += output[i], printf("%lld\n", p);
}