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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
#include <functional>
#include<complex>
using namespace std;

#define rep(i,x) for(ll i=0;i<x;i++)
#define repn(i,x) for(ll i=1;i<=x;i++)

typedef long long ll;
const ll INF = 1e17;
const ll MOD = 1000000007;
const ll MAX = 4000001;
const long double eps = 1E-14;

ll max(ll a, ll b) {
	if (a > b) { return a; }
	return b;
}

ll min(ll a, ll b) {
	if (a > b) { return b; }
	return a;
}

ll gcd(ll a, ll b) {
	if (b == 0) { return a; }
	if (a < b) { return gcd(b, a); }
	return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}

struct edge {
	ll ind;
	ll fr;
	ll to;
	ll d;
};

class mint {
	long long x;
public:
	mint(long long x = 0) : x((x% MOD + MOD) % MOD) {}
	mint operator-() const {
		return mint(-x);
	}
	mint& operator+=(const mint& a) {
		if ((x += a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator-=(const mint& a) {
		if ((x += MOD - a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator*=(const  mint& a) {
		(x *= a.x) %= MOD;
		return *this;
	}
	mint operator+(const mint& a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint& a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint& a) const {
		mint res(*this);
		return res *= a;
	}

	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime MOD
	mint inv() const {
		return pow(MOD - 2);
	}
	mint& operator/=(const mint& a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint& a) const {
		mint res(*this);
		return res /= a;
	}

	friend ostream& operator<<(ostream& os, const mint& m) {
		os << m.x;
		return os;
	}
};

mint pw(mint a, ll b) {
	if (b == 0) { return 1; }
	mint ret = pw(a, b >> 1);
	ret *= ret;
	if (b & 1) { ret *= a; }
	return ret;
}

typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<vector<ll>>> vvvll;

typedef vector<mint> vmint;
typedef vector<vector<mint>> vvmint;
typedef vector<vector<vector<mint>>> vvvmint;

/////////////////////////////////////
ll N, M, K;

vll id;

vll con(vll a,vll b){
	vll ret(N);
	repn(i, N-1)ret[i] = a[b[i]];
	return ret;
}

vll cc(vll a, ll t) {
	if (t == 0) { return id; }
	vll c = cc(a, t / 2);

	vll ret = con(c, c);
	if (t % 2 == 1)ret = con(ret, a);
	return ret;
}

int main() {
	cin >> N;
	vll x(N + 1);
	repn(i, N)cin >> x[i];
	cin >> M >> K;
	vll a(M + 1);
	repn(i, M)cin >> a[i];

	id.resize(N);
	repn(i, N - 1)id[i] = i;

	vll c = id;
	repn(i, M) {
		swap(c[a[i] - 1], c[a[i]]);
	}
	//repn(i, N - 1)cout << c[i]<<" ";
	//cout << endl;

	c = cc(c, K);
	//repn(i, N - 1)cout << c[i] << " ";
	//cout << endl;

	vll d(N);
	repn(i, N - 1) { d[i] = x[i + 1] - x[i]; }
	vll newd(N);
	repn(i, N - 1) { newd[c[i]] = d[i]; }
	//repn(i, N - 1)cout << newd[i] << " ";
	//cout << endl;

	vll ans(N + 1);
	ans[1] = x[1];
	repn(i, N - 1)ans[i + 1] = ans[i] + newd[i];
	repn(i, N)cout << ans[i] << endl;

}