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 <cstdio>
#include <cstdlib>
#include <vector>

using namespace std;

typedef long long ll;
typedef long double ld;

const int N = 1e5 + 500;
const int LOG = 60;

int n, a[N], m, P[N][60];
ll k, x[N], dx[N];

int main(){
	scanf("%d", &n);
	for(int i = 0;i < n;i++)
		scanf("%lld", x + i);
	for(int i = 1;i < n;i++){
		dx[i] = x[i] - x[i - 1];
		P[i][0] = i;
	}	
	scanf("%d%lld", &m, &k);
	for(int i = 0;i < m;i++){
		scanf("%d", a + i), a[i]--;
		swap(P[a[i]][0], P[a[i] + 1][0]);
	}	
	for(int j = 1;j < 60;j++)
		for(int i = 1;i < n;i++){
			P[i][j] = P[P[i][j - 1]][j - 1];
	}
	ll cur = x[0];
	for(int i = 0;i < n;i++){
		int gdje = i;
		for(int j = 0;j < 60;j++)
			if((1LL << j) & k)
				gdje = P[gdje][j];
		cur += dx[gdje];
		printf("%lld\n", cur);
	}
}