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;

typedef pair <int, int> ii;
typedef long long ll;

const int Maxb = 61;
const int Maxn = 100005;

struct arr {
    ii m[4];
    int mlen;
    arr(): mlen(0) {}
};

int n;
int X[Maxn];
int m;
ll k;
arr cur[Maxn], ncur[Maxn];
arr my[Maxb][Maxn];

void Merge(const arr &A, const arr &B, const arr &C, const arr &D, ll multA, ll multB, ll multC, ll multD, arr &E)
{
    int i = 0, j = 0, k = 0, z = 0;
    E.mlen = 0;
    while (i < A.mlen || j < B.mlen || k < C.mlen || z < D.mlen) {
        int mn = Maxn;
        if (i < A.mlen && A.m[i].first < mn) mn = A.m[i].first;
        if (j < B.mlen && B.m[j].first < mn) mn = B.m[j].first;
        if (k < C.mlen && C.m[k].first < mn) mn = C.m[k].first;
        if (z < D.mlen && D.m[z].first < mn) mn = D.m[z].first;
        ll sum = 0;
        if (i < A.mlen && A.m[i].first == mn) { sum += A.m[i].second * multA; i++; }
        if (j < B.mlen && B.m[j].first == mn) { sum += B.m[j].second * multB; j++; }
        if (k < C.mlen && C.m[k].first == mn) { sum += C.m[k].second * multC; k++; }
        if (z < D.mlen && D.m[z].first == mn) { sum += D.m[z].second * multD; z++; }
        if (sum != 0) E.m[E.mlen++] = ii(mn, int(sum));
    }
}

void complexMerge(arr bef[], const arr &cur, arr &fut)
{
    if (cur.mlen == 0) fut.mlen = 0;
    else if (cur.mlen == 1) Merge(bef[cur.m[0].first], arr(), arr(), arr(), cur.m[0].second, 0, 0, 0, fut);
    else if (cur.mlen == 2) Merge(bef[cur.m[0].first], bef[cur.m[1].first], arr(), arr(),
                                  cur.m[0].second, cur.m[1].second, 0, 0, fut);
    else if (cur.mlen == 3) Merge(bef[cur.m[0].first], bef[cur.m[1].first], bef[cur.m[2].first], arr(),
                                  cur.m[0].second, cur.m[1].second, cur.m[2].second, 0, fut);
    else Merge(bef[cur.m[0].first], bef[cur.m[1].first], bef[cur.m[2].first], bef[cur.m[3].first],
               cur.m[0].second, cur.m[1].second, cur.m[2].second, cur.m[3].second, fut);
}

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &X[i]);
        cur[i].mlen = 1;
        cur[i].m[0] = ii(i, 1);
    }
    scanf("%d %lld", &m, &k);
    for (int i = 1; i <= m; i++) {
        int a; scanf("%d", &a);
        arr tmp;
        Merge(cur[a - 1], cur[a], cur[a + 1], arr(), 1, -1, 1, 0, tmp);
        cur[a] = tmp;
    }
    for (int i = 1; i <= n; i++)
        my[0][i] = cur[i];
    for (int b = 1; b < Maxb; b++) {
        for (int i = 1; i <= n; i++)
            complexMerge(my[b - 1], my[b - 1][i], my[b][i]);
    }
    for (int i = 1; i <= n; i++) {
        cur[i].mlen = 1;
        cur[i].m[0] = ii(i, 1);
    }
    for (int b = Maxb - 1; b >= 0; b--) if (1ll << ll(b) <= k) {
        k -= 1ll << ll(b);
        for (int i = 1; i <= n; i++)
            complexMerge(cur, my[b][i], ncur[i]);
        for (int i = 1; i <= n; i++) {
            cur[i] = ncur[i];
            ncur[i].mlen = 0;
        }
    }
    for (int i = 1; i <= n; i++) {
        ll res = 0;
        for (int j = 0; j < cur[i].mlen; j++)
            res += ll(X[cur[i].m[j].first]) * cur[i].m[j].second;
        printf("%lld.0\n", res);
    }
    return 0;
}