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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
const ll maxn = 100000 + 10;

ll n, m, a[maxn], b[maxn], vis[maxn], d[maxn], ans[maxn];
ll k;

int main() {
  scanf("%lld", &n);
  for(ll i = 1, sum = 0;i <= n;i ++) {
    scanf("%lld", &a[i]);
    a[i] -= sum;
    sum += a[i];
    b[i] = i;
  }
  scanf("%lld%lld", &m, &k);
  for(ll i = 1;i <= m;i ++) {
    ll x; scanf("%lld", &x);
    swap(b[x],b[x + 1]);
  }
  for(ll i = 1, tot;i <= n;i ++) {
    if(!vis[i]) {
      tot = 0;
      for(ll j = i;!vis[j];j = b[j]) {
      	vis[j] = 1;
      	d[++ tot] = j;
      }
      for(ll j = 1;j <= tot;j ++) {
        ans[d[j]] = a[d[(j + k - 1) % tot + 1]];
	  }
	}
  }
  for(ll i = 1;i <= n;i ++) {
    ans[i] += ans[i - 1];
    printf("%lld.0\n", ans[i]);
  }
  return 0;
}