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 <stdio.h>
int n, m, a, b, c, d;
long long x[100000];
long long k;
int s[100000];
int t[100000];
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%lld", x + i);
		s[i] = i;
	}
	for (int i = n - 1; i; i--)x[i] -= x[i - 1];
	scanf("%d%lld", &m, &k);
	for (int i = 0; i < m; i++) {
		scanf("%d", &a);
		b = s[a - 1];
		s[a - 1] = s[a];
		s[a] = b;
	}
	for (int i = 1; i < n; i++) {
		if (t[i])continue;
		c = s[i];
		d = 1;
		while (c != i) {
			c = s[c];
			d++;
		}
		b = k % d;
		c = i;
		for (int j = 0; j < b; j++) {
			c = s[c];
		}
		b = i;
		for (int j = 0; j < d; j++) {
			t[b] = c;
			b = s[b];
			c = s[c];
		}
	}
	k = 0;
	for (int i = 0; i < n; i++) {
		k += x[t[i]];
		printf("%lld\n", k);
	}
}