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
#include<bits/stdc++.h>
using namespace std;
#define db double 
#define LL long long 
const int N = 1e5 + 11;
db x[N];
int m, a[N], n;
int stk[N], tp, to[N], res[N];
bool mark[N];
LL K;
int main(){
    cin>>n;
    for(int i = 1;i <= n; i++){
        scanf("%lf", &x[i]);
    }
    for(int i = n;i >= 1; i--)x[i] -= x[i-1], to[i] = i;
    cin>>m>>K;
    for(int i = 1;i <= m; i++){
        scanf("%d", &a[i]);
        swap(to[a[i]], to[a[i]+1]);
    }
    for(int i = 1;i <= n; i++){
        if(mark[i])continue;
        mark[i] = tp = 1; stk[1] = i; 
        int now = to[i];
        while(now != i)mark[now] = 1, stk[++tp] = now, now = to[now];
        for(int j = 1;j <= tp; j++)res[stk[j]] = stk[(j-1+K)%tp+1];
    }
    db t = 0;
    for(int i = 1;i <= n; i++){
        t += x[res[i]];
        printf("%.1lf\n", t);
    }
    return 0;
}
