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 <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define pi acos(-1.0)
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
 
int main(){
    int n, m;
    ll k;
    cin >> n;
    VI x(n);
    REP(i,n) scanf("%d", &x[i]);
    cin >> m >> k;
    VI y(n-1);
    REP(i,n-1) y[i] = x[i+1] - x[i];
    VI s(n-1), a(n-1);
    REP(i,n-1) a[i] = s[i] = i;
    REP(i,m){
        int d;
        scanf("%d", &d);
        d--;
        swap(s[d-1], s[d]);
    }

    VI so(s);
    REP(i,n-1) s[so[i]] = i;
 
    while (k){
        if (k & 1LL){
        	REP(i,n-1) a[i] = s[a[i]];
        }
        VI so(s);
        REP(i,n-1) s[i] = so[s[i]];
        k /= 2;
    }

    VI ao(a);
    REP(i,n-1) a[ao[i]] = i;

    VI e(y);
    REP(i,n-1) y[a[i]] = e[i];
 
    int ans = x[0];
    cout << ans << endl;
    REP(i,n-1){
        ans += y[i];
        printf("%d\n",ans);
    }
 
    return 0;
}