Problem:
Score : 800 points

Problem Statement

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}.

Sample Input 1

3
-1 0 2
1 1
2

Sample Output 1

-1.0
1.0
2.0

Rabbit 2 will perform the jump.
If rabbit 1 is chosen, the coordinate of the destination will be -2.
If rabbit 3 is chosen, the coordinate of the destination will be 4.
Thus, the expected value of the coordinate of the eventual position of rabbit 2 is 0.5×(-2)+0.5×4=1.0.

Sample Input 2

3
1 -1 1
2 2
2 2

Sample Output 2

1.0
-1.0
1.0

x_i may not be distinct.

Sample Input 3

5
0 1 3 6 10
3 10
2 3 4

Sample Output 3

0.0
3.0
7.0
8.0
10.0

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//{{{
#include <bits/stdc++.h>
using namespace std;
#define SZ(x) ((int)(x).size())
#define ALL(x) x.begin(), x.end()
#define REP(i, n) for (int i = 0; i < int(n); ++i)
#define REPE(i, a, b) for (int i = (a); i <= int(b); ++i)
#define DB(x) cerr << __LINE__ << ": " << #x << " = " << (x) << endl
#define MP make_pair
#define PB push_back
#define EB emplace_back
#define BR putchar('\n')
#define MS0(ar) memset(ar, 0, sizeof(ar))
#define MS1(ar) memset(ar, -1, sizeof(ar))
#define F first
#define S second
#define MP make_pair
#define MT make_tuple
#define IO(name) freopen(name".in", "r", stdin);freopen(name".out", "w", stdout)

typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
typedef pair<int, int> PII;
typedef complex<double> CD;
typedef vector<int> VI;
typedef vector<double> VD;
typedef vector<LL> VL;

void R(int &x) { scanf("%d", &x); }
void R(LL &x) { scanf("%lld", &x); }
void R(ULL &x) { scanf("%llu", &x); }
void R(double &x) { scanf("%lf", &x); }
void R(LD &x) { scanf("%Lf", &x); }

template <typename T> void R(T &t) { cin >> t; }
template <typename A, typename B> void R(pair<A, B> &x) {
  R(x.F);
  R(x.S);
}

template <typename T> void R(vector<T> &ar) {
  for (auto &it : ar)
    R(it);
}

template <typename T, typename... Args> void R(T &t, Args &... args) {
  R(t);
  R(args...);
}

void W(const char &c) { putchar(c); };
void W(const int &x) { printf("%d", x); }
void W(const LL &x) { printf("%lld", x); }
void W(const ULL &x) { printf("%llu", x); }
void W(const double &x) { printf("%lf", x); }
void W(const LD &x) { printf("%Lf", x); }

template <typename T> void W(const T &t) { cout << t; }
template <typename T> void W(const vector<T> &ar) {
  for (size_t i = 0; i < ar.size(); ++i) {
    W(ar[i]);
    putchar(" \n"[i + 1u == ar.size()]);
  }
}
template <typename T, typename... Args>
void W(const T &t, const Args &... args) {
  W(t);
  W(args...);
}
#define BR putchar('\n')

template <typename T> void Min(T &x, T v) { x = min(x, v); }
template <typename T> void Max(T &x, T v) { x = max(x, v); }
//}}}
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
const int maxn = 100010;

int n, c[maxn];
double a[maxn];

int m;
long long k;

int main() {
  R(n);
  REP(i, n) R(a[i]);
  R(m, k);
  if (k > 800) {
    k = k % 400 + 400;
  }
  REP(i, m) {
    R(c[i]);
    c[i]--;
  }
  while (k--) {
    REP(i, n) {
      int x = c[i];
      a[x] = a[x - 1] + a[x + 1] - a[x];
    }
  }
  REP(i, n) printf("%.12f\n", a[i]);
  return 0;
}