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>
//#include<tr1/unordered_map>
//#include"Bignum/bignum.h"
//#define lll bignum
#define ls(x) ( x << 1 )
#define rs(x) ( x << 1 | 1 )
//#define mid ( ( l + r ) >> 1 )
#define lowbit(x) ( x & -x )
#define debug(x) (cout << "#x = " << x << endl)
#define re register
#define For(i, j, k) for(re int i = (j); i <= (k); i++)
#define foR(i, j, k) for(re int i = (j); i >= (k); i--)
#define Cross(i, j, k) for(re int i = (j); i; i = (k))
using namespace std;
typedef long long ll;
const ll N = 100011;
const ll p = 998244353;
const ll inf = 0x3f3f3f3f3f3f;

ll n, m, k, x, res = 0, a[N], b[N], c[N], B[N];

namespace IO {

	#define dd ch = getchar()
	inline ll read() {
		ll x = 0; bool f = 0; char dd;
		for (; !isdigit (ch); dd) f ^= (ch == '-');
		for (; isdigit (ch); dd)  x = (x << 3) + (x << 1) + (ch ^ 48);
		return f? -x: x;
	}
	#undef dd

	inline void write( ll x ) {
		if ( x < 0 ) putchar ( '-' ), x = -x;
		if ( x > 9 ) write ( x / 10 ); putchar ( x % 10 | 48 );
	}

	inline void wrn ( ll x ) { write (x); putchar ( ' ' ); }

	inline void wln ( ll x ) { write (x); putchar ( '\n' ); }

	inline void wlnn ( ll x, ll y ) { wrn (x), wln (y); }

}

using namespace IO;

inline void Redouble ( ll x ) {
	for (; x; x >>= 1) {
		if (x & 1) {
			For ( i, 1, n ) B[i] = a[b[i]]; 
			For ( i, 1, n ) a[i] = B[i];
		}
		For ( i, 1, n ) B[i] = b[b[i]]; 
		For ( i, 1, n ) b[i] = B[i];
	}
}

int main() 
{
//	freopen("qwq.in", "r", stdin);
//	freopen("qwq.out", "w", stdout);
	n = read(); 
	For ( i, 1, n ) a[i] = read(), b[i] = i;
	foR ( i, n, 1 ) a[i] = a[i] - a[i - 1];
	m = read(), k = read();
	For ( i, 1, m ) x = read(), swap ( b[x], b[x + 1] );
	Redouble (k);
	For ( i, 1, n ) 
		res = (res + a[i] + p) % p, wln (res);
	return 0;
}

/*

*/