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>
#define LL long long
using namespace std;
const int N = 1e5 + 7;

LL a[N], b[N], c[N];
int f[N];
bool vis[N];

int main(){
	int n;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
	for (int i = 2; i <= n; i++) b[i] = a[i] - a[i - 1];
	for (int i = 2; i <= n; i++) f[i] = i;
	int m; long long k;
	scanf("%d%lld", &m, &k);
	for (int i = 1; i <= m; i++){
		int x; scanf("%d", &x);
		swap(f[x], f[x + 1]);
	}
	for (int i = 2; i <= n; i++){
		if (!vis[i]){
			vector<int > v;
			int j = i;
			while (!vis[j]){
				v.push_back(j);
				vis[j] = true;
				j = f[j];
			}	
			int len = v.size(), p = k % len;
			for (int j = 0; j < len; j++){
				c[v[j]] = b[v[(j + p) % len]];
			}
		}
	}
	LL t = a[1];
	for (int i = 2; i <= n; i++){
		t += c[i];
		a[i] = t;
	}
	for (int i = 1; i <= n; i++) printf("%lld ", a[i]);

	return 0;
}