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 <bits/stdc++.h>
using namespace std;
#define MAXN 100010
int n,m,a[MAXN],ans[MAXN],res[MAXN],tmp[MAXN];
long long k;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;++i)scanf("%d",&a[i]);
    for(int i=1;i<n;++i)ans[i]=i;
    scanf("%d%lld",&m,&k);
    for(int i=1;i<=m;++i){
        int x;
        scanf("%d",&x);
        swap(ans[x],ans[x-1]);
    }
    for(int i=1;i<n;++i)res[i]=i;
    while(k){
        if(k&1)
            for(int i=1;i<n;++i)res[i]=ans[res[i]];
        for(int i=1;i<n;++i)tmp[i]=ans[ans[i]];
        for(int i=1;i<n;++i)ans[i]=tmp[i];
        k>>=1;
    }
    long long sum=a[1];
    for(int i=1;i<=n;++i){
        printf("%lld.0\n",sum);
        sum+=a[res[i]+1]-a[res[i]];
    }
}