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;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int n, x[100001], p[100000], d[100000], np[100000], tp[100000];
ll an[100001];

int main(){
	scanf("%d", &n);
	f(i, 1, n + 1)scanf("%d", x + i);
	f(i, 1, n)d[i] = x[i + 1] - x[i];
	f(i, 1, n)p[i] = i, np[i] = i;
	int m;
	ll k;
	scanf("%d%lld", &m, &k);
	f(i, 0, m){
		int t;
		scanf("%d", &t);
		swap(p[t - 1], p[t]);
	}
	while (k){
		if (k & 1){
			f(i, 1, n)tp[i] = np[p[i]];
			f(i, 1, n)np[i] = tp[i];
		}
		f(i, 1, n)tp[i] = p[p[i]];
		f(i, 1, n)p[i] = tp[i];
		k >>= 1;
	}
	an[1] = x[1];
	f(i, 2, n + 1)an[i] = an[i - 1] + d[np[i - 1]];
	f(i, 1, n + 1)printf("%lld\n", an[i]);
}
