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 ll long long
#define MAX 100100

int n, x[MAX], m, d[MAX], ans[MAX], tmp[MAX];
ll K;
double a[MAX];

int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &x[i]);
  scanf("%d%lld", &m, &K);
  for (int i = 1; i <= n; ++i) d[i] = i, ans[i] = i;
  for (int i = 1, x; i <= m; ++i) scanf("%d", &x), swap(d[x], d[x + 1]);
  while (K) {
    if (K & 1) {
      for (int i = 1; i <= n; ++i) tmp[i] = ans[d[i]];
      for (int i = 1; i <= n; ++i) ans[i] = tmp[i];
    }
    for (int i = 1; i <= n; ++i) tmp[i] = d[d[i]];
    for (int i = 1; i <= n; ++i) d[i] = tmp[i];
    K >>= 1;
  }
  for (int i = 1; i <= n; ++i) a[i] = x[ans[i]] - x[ans[i] - 1];
  for (int i = 1; i <= n; ++i) printf("%.1lf\n", a[i] += a[i - 1]);
  return 0;
}