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<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#define ll long long
using namespace std;

const int maxn=1e5+15;
ll n,m,k,cnt;
ll vis[maxn],s[maxn],x[maxn],num[maxn],be[maxn],p[maxn],a[maxn];
vector <ll> q[maxn];
void dfs(ll x,ll dep)//找环 
{
	num[x]=dep;vis[x]=1;be[x]=cnt;
	q[cnt].push_back(x);
	if (!vis[p[x]]) dfs(p[x],dep+1);
}
int main()
{
	scanf("%lld",&n);
	for (int i=1;i<=n;i++) scanf("%lld",x+i);
	for (int i=1;i<n;i++) s[i]=x[i+1]-x[i];//差分序列 
	scanf("%lld%lld",&m,&k);
	for (int i=1;i<n;i++) p[i]=i;
	for (int i=1;i<=m;i++) {scanf("%lld",a+i);swap(p[a[i]],p[a[i]-1]);}//一次置换的结果，记录下一次置换兔子到哪了 
	for (int i=1;i<n;i++) if (!vis[i]) {dfs(i,0);++cnt;}
	ll now=x[1];
	printf("%lld\n",now);
	for (int i=1;i<n;i++)
	{
		now+=s[q[be[i]][(num[i]+k)%q[be[i]].size()]];
		printf("%lld\n",now);
	}
	return 0;
}