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"
#include<cassert>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
const int mod = 1000000007;
const int inf = 1ll << 61;
typedef pair<int, int> P;
typedef pair<P,int>PP;
typedef vector<vector<int>>mat;

int amida[63][100006];

int x[100006];
int a[100006];
int fin[100006];
signed main() {
	int n; cin >> n;
	rep(i, n)cin >> x[i];
	int m, k; cin >> m >> k;
	rep(i, m) {
		cin >> a[i]; a[i]--;
	}

	rep(i, n)amida[0][i] = i;
	rep(i,m) {
		swap(amida[0][a[i]], amida[0][a[i] - 1]);
	}
	for (int i = 0; i < 62; i++) {
		rep(j, n) {
			amida[i + 1][j] = amida[i][amida[i][j]];
		}
	}
	rep(i, n)fin[i] = i;
	rep(i, 63) {
		if (k&(1ll << i)) {
			rep(j, n) {
				fin[j] = amida[i][fin[j]];
			}
		}
	}
	int cnt = x[0];
	cout << cnt<<endl;
	rep(i, n - 1) {
		cnt += x[fin[i] + 1] - x[fin[i]];
		cout << cnt << endl;
	}
}