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;
typedef long long ll;
const int N = 123455, LG = 60;
ll n, m, k, A[N], P[N], R[LG][N];
int main()
{
    scanf("%lld", &n);
    for (int i = 0; i < n; i++)
        scanf("%lld", &A[i]), P[i] = i;
    scanf("%lld%lld", &m, &k);
    for (int i = 0; i < m; i++)
    {
        int a;
        scanf("%lld", &a);
        swap(P[a], P[a - 1]);
    }
    for (int i = 0; i < n; i++)
        R[0][i] = P[i];
    for (int j = 1; j < LG; j++)
        for (int i = 0; i < n; i++)
            R[j][i] = R[j - 1][R[j - 1][i]];
    printf("%lld\n", A[0]); ll nw = A[0];
    for (int i = 1; i < n; i++)
    {
        int id = i;
        for (int j = 0; j < LG; j++)
            if ((k >> j) & 1LL)
                id = R[j][id];
        nw += A[id] - A[id - 1];
        printf("%lld\n", nw);
    }
    return 0;
}