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<algorithm>
using namespace std;
const int MAXN = 100000;
typedef long long ll;
ll x[MAXN + 5], d[MAXN + 5];
int a[MAXN + 5]; bool tag[MAXN + 5];
int b[MAXN + 5], ans[MAXN + 5];
int main() {
	int N, M; ll K;
	scanf("%d", &N);
	for(int i=1;i<=N;i++) {
		scanf("%lld", &x[i]);
		if( i != 1 ) d[i-1] = x[i] - x[i-1];
	}
	scanf("%d%lld", &M, &K);
	for(int i=1;i<N;i++)
		a[i] = i;
	for(int i=1;i<=M;i++) {
		int x;
		scanf("%d", &x);
		swap(a[x-1], a[x]);
	}
	for(int i=1;i<N;i++) {
		if( tag[i] ) continue;
		int p = i, siz = 0;
		do {
			tag[p] = true;
			b[siz++] = p;
			p = a[p];
		}while( p != i );
		int q = 0;
		for(int j=0;j<siz;j++) {
			ans[b[j]] = b[(j+K)%siz];
		}
	}
	ll nw = x[1];
	for(int i=1;i<=N;i++) {
		printf("%lld\n", nw);
		if( i != N ) nw += d[ans[i]];
	}
}