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;
using ll = long long;

// #undef DEBUG
// #define DEBUG
/// {{{ DEBUG --- ///
template <typename T> ostream &operator<<(ostream &o, const vector<T> &v) { if(&o == &cerr) { o << '{'; for(size_t i = 0; i < v.size(); i++) o << v[i] << (i + 1 != v.size() ? ", " : ""); o << "}"; } else { for(size_t i = 0; i < v.size(); i++) o << v[i] << (i + 1 != v.size() ? " " : ""); } return o; }
#ifdef DEBUG
#ifdef USE_COUT
#define dump(...) (cout<<"["<<__LINE__<< "] "<<#__VA_ARGS__<<" = "<<make_tuple(__VA_ARGS__)<<"\n")
#else
#define dump(...) (cerr<<"["<<__LINE__<< "] "<<#__VA_ARGS__<<" = "<<make_tuple(__VA_ARGS__)<<"\n")
#endif
template<int n, class...T> typename enable_if<(n>=sizeof...(T))>::type _ot(ostream &, tuple<T...> const &){}
template<int n, class...T> typename enable_if<(n< sizeof...(T))>::type _ot(ostream & os, tuple<T...> const & t){ os << (n==0?"":", ") << get<n>(t); _ot<n+1>(os, t); }
template<class...T> ostream & operator<<(ostream &o, tuple<T...> const &t){ o << "("; _ot<0>(o, t); o << ")"; return o; }
template<class T, class U> ostream & operator<<(ostream &o, pair<T, U> const &p) { o << "(" << p.first << ", " << p.second << ")"; return o; }
#else
#define dump(...) (0)
#endif
/// }}}--- ///

const int N = 1e5;
int n;
ll m, k;
ll a[N];
ll d[N];
int o[N];

vector<int> mult(vector<int> &a, vector<int> &b){ 
  auto res = a;
  for(int i = 0; i < n; i++) {
    res[b[a[i]]] = i;
  }
  return res;
}

int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for(int i = 0; i < n; i++) {
    cin >> a[i];
    if(i) d[i] = a[i] - a[i - 1];
  }
  cin >> m >> k;
  vector<int> ord(n);
  iota(begin(ord), end(ord), 0);
  for(int i = 0; i < m; i++) {
    cin >> o[i], o[i]--;
    swap(ord[o[i]], ord[o[i] + 1]);
  }
  vector<int> ans(n);
  iota(begin(ans), end(ans), 0);
  while(k) {
    if(k & 1) ans = mult(ans, ord);
    ord = mult(ord, ord);
    k >>= 1;
  }
  vector<ll> x(n);
  for(int i = 0; i < n; i++) {
    x[ans[i]] = d[i];
  }
  for(int i = 1; i < n; i++) a[i] = a[i - 1] + x[i];
  for(int i = 0 ;i < n; i++) cout << a[i] << endl;
}