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 fi first
#define se second
typedef pair<int,int> PII;


long long n, m, k, a[1<<20], ans[1<<20], d[1<<20], p[1<<20];

void mul(long long a[], long long b[]) {
    static long long c[1<<20];
    for (int i = 1; i < n; i++) c[i] = a[b[i]];
    for (int i = 1; i < n; i++) a[i] = c[i];
}

int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        d[i+1] -= a[i], d[i] += a[i];
        ans[i+1] = p[i+1] = i+1;
    }
    cin >> m >> k;
    for (int i = 0; i < m; i++) {
        int u;
        cin >> u;
        swap(p[u],p[u-1]);
    }
    for (long long x = k; x; x>>=1) { if (x&1) mul(ans,p); mul(p,p); }
    for (int i = 1; i < n; i++) a[i] = a[i-1] + d[ans[i]];
    for (int i = 0; i < n; i++) cout << a[i] << '\n';
}
