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>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define svec(v) cout<<#v<<":";rep(kbrni,v.size())cout<<" "<<v[kbrni];cout<<endl
#define sset(s) cout<<#s<<":";each(kbrni,s)cout<<" "<<kbrni;cout<<endl
#define smap(m) cout<<#m<<":";each(kbrni,m)cout<<" {"<<kbrni.first<<":"<<kbrni.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<double> vd;
typedef vector<P> vp;

const int MAX_N = 100005;

int dp[MAX_N][60];

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<double> x(n),y(n-1);
    rep(i,n){
        cin >> x[i];
    }
    rep(i,n-1){
        y[i] = x[i+1] - x[i];
    }
    vector<int> vec(n-1);
    rep(i,n-1){
        vec[i] = i;
    }
    int m;
    ll K;
    cin >> m >> K;
    rep(i,m){
        int a;
        cin >> a;
        swap(vec[a-2],vec[a-1]);
    }
    rep(i,n-1){
        dp[i][0] = vec[i];
    }
    rep(i,58){
        rep(j,n-1){
            dp[j][i+1] = dp[dp[j][i]][i];
        }
    }
    rep(i,n-1){
        vec[i] = i;
    }
    int keta = 0;
    vector<int> tmp(n-1);
    rep(keta,58){
        if(K & (1LL<<keta)){
            rep(i,n-1){
                tmp[i] = dp[vec[i]][keta];
            }
            rep(i,n-1){
                vec[i] = tmp[i];
            }
        }
    }
    zip(vec);
    if(len(vec) < n-1){
        assert(false);
    }
    printf("%f\n",x[0]);
    rep(i,n-1){
        x[i+1] = x[i]+y[vec[i]];
        printf("%f\n",x[i+1]);
    }
    return 0;
}