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>
#pragma GCC optimize("Ofast", "unroll-loops")
#define ll long long 
#define pii pair<int, int>
#define ull unsigned ll
#define f first
#define s second
#define FOR(i,a,b) for (int i=(a); i<(b); i++)
#define REP(i,n) for (int i=0; i<(n); i++)
#define RREP(i,n) for (int i=(n-1); i>=0; i--)
#define ALL(x) x.begin(),x.end()
#define SZ(x) x.size()
#define SQ(x) (x)*(x)
#define MNTO(a,b) a = min(a,(__typeof__(a))(b))
#define MXTO(a,b) a = max(a,(__typeof__(a))(b))
#define pb push_back
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define debug(x) cerr<<#x<<" is "<<x<<endl
using namespace std;

// #define int ll

const int iinf = 1<<29;
const ll inf = 1ll<<60;
const ll mod = 1e9+7;


void GG(){cout<<"No\n"; exit(0);}

ll mpow(ll a, ll n){ // a^n % mod
    ll re=1;
    while (n>0){
        if (n&1) re = re*a %mod;
        a = a*a %mod;
        n>>=1;
    }
    return re;
}

ll inv (ll b){
    if (b==1) return b;
    return (mod-mod/b) * inv(mod%b) % mod;
}

const int maxn = 1e5+5;

vector<int> p;
int pos[maxn];

bool done[maxn];

main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    ll n; cin>>n;
    vector<ll> x(n); REP(i,n) cin>>x[i];
    ll m, k; cin>>m>>k;
    vector<ll> v(n);
    vector<ll> d;
    REP(i,m){
        cin>>v[i];
    }
    REP(i,n){
        if (i) d.pb(x[i] - x[i-1]);
    }
    // k%=2;
    k--;
    REP(i,n-1) p.pb(i);
    REP(i,m){
        swap(p[v[i]-2], p[v[i]-1]);
    }
    REP(i,n-1){
        if (done[i]) continue;
        vector<int> tmp;
        int at = i; 
        tmp.pb(at); at = p[at];
        while (at != i){
            tmp.pb(at); at = p[at];
        }
        int j = tmp.size(); int df = k%j;
        FOR(k,i,i+j){
            pos[k]=(p[k] + df );
            if (pos[k] >= i+j) pos[k] -= j;

            done[k] = 1;
        }
    }

    ll at = x[0];
    REP(i,n){
        cout<<at<<'\n';
        if (i!=n-1) at += d[pos[i]];
    }
}