Problem: 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}.

Examples

Input

3
-1 0 2
1 1
2


Output

-1.0
1.0
2.0


Input

3
1 -1 1
2 2
2 2


Output

1.0
-1.0
1.0


Input

5
0 1 3 6 10
3 10
2 3 4


Output

0.0
3.0
7.0
8.0
10.0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

typedef long double ld;
typedef long long ll;

const ll N = 1e5 + 10;

ll n, m; ll k;
ll x[N], a[N], b[N], c[N];
ll st[N][61];

int main() {
  scanf("%lld", &n);
  for (ll i = 1; i <= n; i++) {
    b[i] = i;
    scanf("%lld", &x[i]);
  }
  scanf("%lld %lld", &m, &k);
  for (ll i = 1; i <= m; i++) {
    scanf("%lld", &a[i]);
    swap(b[a[i] - 1], b[a[i]]);
  }
  for (ll i = 1; i < n; i++) st[b[i]][0] = i;
  for (ll j = 1; j < 61; j++)
    for (ll i = 1; i < n; i++)
      st[i][j] = st[st[i][j - 1]][j - 1];
  for (ll i = 1; i < n; i++) {
    ll cur = i;
    for (ll j = 0; j < 61; j++) {
      if ((k >> j) & 1) {
        cur = st[cur][j];
      }
    }
    c[cur + 1] = x[i + 1] - x[i];
  }
  c[1] = x[1];
  for (ll i = 2; i <= n; i++) c[i] += c[i - 1];
  for (ll i = 1; i <= n; i++) printf("%lld\n", c[i]);
  return 0;
}