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 <cstring>
#include <algorithm>
 
using namespace std;
 
typedef long long ll;
 
int num[100005],s[100005];
int id[100005],now[100005];
bool vis[100005];
int ans[100005];
 
int main() {
  int n;
  scanf("%d",&n);
  for(int i=1;i<=n;i++) {
    scanf("%d",&num[i]);
    if (i>1) s[i-1]=num[i]-num[i-1];
  }
  for(int i=1;i<n;i++) id[i]=i;
  int m;
  ll k=0;
  scanf("%d%lld",&m,&k);
  for(int i=1;i<=m;i++) {
    int x;
    scanf("%d",&x);
    swap(id[x-1],id[x]);
  }
  for(int i=1;i<n;i++)
    if (!vis[i]) {
      int x=i,cnt=0;
      while (!vis[x]) {
        now[cnt++]=x;
        vis[x]=1;
        x=id[x];
      }
      for(int i=0;i<cnt;i++)
        ans[now[i]]=now[(i+k)%cnt];
    }
  ll sum=num[1];
  for(int i=1;i<=n;i++) {
    printf("%lld.0\n",sum);
    sum+=s[ans[i]];
  }
  return 0;
}