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>
#include<algorithm>
using namespace std;
int n, w[101000], P[101000], m, TP[101000], R[101000];
long long K, bb;
void Do(int *a, int *b){
    int i;
    for(i=1;i<=n;i++){
        TP[i] = a[b[i]];
    }
    for(i=1;i<=n;i++)a[i] = TP[i];
}
int main(){
    int i, a;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%d",&w[i]);
    }
    bb = w[1];
    n--;
    for(i=1;i<=n;i++)w[i] = w[i+1]-w[i];
    for(i=1;i<=n;i++)P[i] = R[i] = i;
    scanf("%d%lld",&m,&K);
    while(m--){
        scanf("%d",&a);
        swap(P[a],P[a-1]);
    }
    while(K){
        if(K%2){
            Do(R,P);
        }
        Do(P,P);
        K/=2;
    }
    printf("%lld\n",bb);
    for(i=1;i<=n;i++){
        bb += w[R[i]];
        printf("%lld\n",bb);
    }
}