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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;

using ll=long long;
vector<int> merge(vector<int> lhs,vector<int> rhs){
    int n=lhs.size();
    vector<int> res(n);
    for(int i=0;i<n;i++){
        res[i]=lhs[rhs[i]];
    }
    return res;
}
int main(){
    int n;
    cin>>n;
    vector<ll> x(n);
    for(int i=0;i<n;i++) cin>>x[i];
    int m; ll k;
    cin>>m>>k;   
    vector<int> a(n);
    for(int i=0;i<m;i++) cin>>a[i],a[i]--;

    // cerr<<"#PERM PHASE"<<endl;
    vector<int> perm(n-1);
    iota(perm.begin(),perm.end(),0);
    for(int i=0;i<m;i++){
        // assert(0<=a[i]-1 && a[i]<n-1);
        swap(perm[a[i]-1],perm[a[i]]);
    }
    vector<int> diff(n-1);
    for(int i=0;i+1<n;i++) diff[i]=x[i+1]-x[i];

    // cerr<<"#POW PHASE"<<endl;
    while(k){
        if(k&1) diff=merge(diff,perm);
        k>>=1;
        perm=merge(perm,perm);
    }

    // cerr<<"#RESTORE PHASE"<<endl;
    vector<ll> res(n);
    res[0]=x[0];
    for(int i=1;i<n;i++){
        res[i]=res[i-1]+diff[i-1];
    }
    for(int i=0;i<n;i++){
        cout<<res[i]<<endl;
    }
    return 0;
}