Problem: Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.

You have an array a of length 2n and m queries on it. The i-th query is described by an integer qi. In order to perform the i-th query you must:

  * split the array into 2n - qi parts, where each part is a subarray consisting of 2qi numbers; the j-th subarray (1 ≤ j ≤ 2n - qi) should contain the elements a[(j - 1)·2qi + 1], a[(j - 1)·2qi + 2], ..., a[(j - 1)·2qi + 2qi]; 
  * reverse each of the subarrays; 
  * join them into a single array in the same order (this array becomes new array a); 
  * output the number of inversions in the new a. 



Given initial array a and all the queries. Answer all the queries. Please, note that the changes from some query is saved for further queries.

Input

The first line of input contains a single integer n (0 ≤ n ≤ 20). 

The second line of input contains 2n space-separated integers a[1], a[2], ..., a[2n] (1 ≤ a[i] ≤ 109), the initial array.

The third line of input contains a single integer m (1 ≤ m ≤ 106). 

The fourth line of input contains m space-separated integers q1, q2, ..., qm (0 ≤ qi ≤ n), the queries.

Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.

Output

Output m lines. In the i-th line print the answer (the number of inversions) for the i-th query.

Examples

Input

2
2 1 4 3
4
1 2 0 2


Output

0
6
6
0


Input

1
1 2
3
0 1 1


Output

0
1
0

Note

If we reverse an array x[1], x[2], ..., x[n] it becomes new array y[1], y[2], ..., y[n], where y[i] = x[n - i + 1] for each i.

The number of inversions of an array x[1], x[2], ..., x[n] is the number of pairs of indices i, j such that: i < j and x[i] > x[j].

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

@SuppressWarnings("unused")
public class Solution{
	
	static int p, n;
	static int[] a;
	static long[] equals;  // no. of equal pairs in p power
	static HashMap<Integer, ArrayList<Integer>> map;
	
	
	@SuppressWarnings("unchecked")
	public static void main(String[] args) throws IOException {
		FastScanner fs = new FastScanner();
    	PrintWriter out = new PrintWriter(System.out);
    	
    	int tt = 1;
    	outer:
    	while(tt-->0) {
    		
    		p = fs.nextInt();
    		n = (1<<p);
    		
    		map = new HashMap<Integer, ArrayList<Integer>>();
    		equals = new long[p+1];
    		long[] EQUALS = equals;
    		
    		a = new int[n+1];
    		for(int i=1;i<=n;i++) {
    			a[i] = fs.nextInt();
    			if(!map.containsKey(a[i])) map.put(a[i], new ArrayList<Integer>());
    			map.get(a[i]).add(i);
    		}
    		
    		int m = fs.nextInt();
    		
    		long[] val = new long[p+1];
    		val[0] = 0;
    		
    		for(int i=1;i<=p;i++) {
    			val[i] += val[i-1];
    			for(int j=1;j<=n;j+=(1<<i)) {
    				int l = j, r = j + (1<<i) - 1;
    				val[i] += mergeCount(a, l, (l+r)/2, r, i);
    			}
    		}
    		
    		for(int i=1;i<=p;i++) equals[i] += equals[i-1];
    		
    		long[] newval = new long[p+1];
    		for(int i=0;i<=p;i++) newval[i] = val[i];
    		
    		while(m-->0) {
    			int q = fs.nextInt();
    			for(int i=0;i<=p;i++) {
    				if(i<=q) {
    					newval[i] += C2(1<<i, i) - 2*val[i]; 
    				}
    				else {
    					newval[i] += C2(1<<q, q) - 2*val[q];
    				}
    			}
    			for(int i=0;i<=p;i++) val[i] = newval[i];
    			out.println(val[p]);
    		}
    		
    		
    		
    	}
    	
    	out.close();
    }
	
	//total no. of non equal pairs
	static long C2(int x, int k) {
		return (long)(1<<(p-k))*x*(x-1)/2 - equals[k];
	}
	
	//merges two arrays and count inversions between them
	static long mergeCount(int[] arr, int l, int m, int r, int p) {
		long num = 0;
		int n = (r-l+1)/2;
		int[] left = Arrays.copyOfRange(arr, l, m+1), right = Arrays.copyOfRange(arr, m+1, r+1);
		int ind1 = 0, ind2 = 0;
		
		HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
		for(int i=l;i<=m;i++) map.put(arr[i], map.getOrDefault(arr[i], 0)+1);
		
		for(int i=l;i<=r;i++) {
			if(ind1<n && (ind2==n || left[ind1]<=right[ind2])) {
				arr[i] = left[ind1]; ind1++;
			}
			else {
				equals[p] += map.getOrDefault(right[ind2], 0);
				arr[i] = right[ind2]; ind2++;
				num += n-ind1;
			}
		}
		return num;
	}
	
	

	static final Random random=new Random();
    
    static <T> void shuffle(T[] arr) {
    	int n = arr.length;
    	for(int i=0;i<n;i++	) {
    		int k = random.nextInt(n);
    		T temp = arr[k]; arr[k] = arr[i]; arr[i] = temp;
    	}
    }
    
    	
    static void ruffleSort(int[] a) {
    	int n=a.length;//shuffle, then sort 
    	for (int i=0; i<n; i++) {
    		int oi=random.nextInt(n); int temp=a[oi];
    		a[oi]=a[i]; a[i]=temp;
    	}
    	Arrays.sort(a);
    }
    
    static void ruffleSort(long[] a) {
    	int n=a.length;//shuffle, then sort 
    	for (int i=0; i<n; i++) {
    		int oi=random.nextInt(n); long temp=a[oi];
    		a[oi]=a[i]; a[i]=temp;
    	}
    	Arrays.sort(a);
    }
   
  
    
    static void reverse(int[] arr, int l, int r) {
    	for(int i=l;i<l+(r-l)/2;i++){
    		int temp = arr[i]; arr[i] = arr[r-i+l-1]; arr[r-i+l-1] = temp;
    	}
    }
    
    static void reverse(long[] arr, int l, int r) {
    	for(int i=l;i<l+(r-l)/2;i++){
    		long temp = arr[i]; arr[i] = arr[r-i+l-1]; arr[r-i+l-1] = temp;
    	}
    }
    
    
    static <T> void reverse(T[] arr, int l, int r) {
    	for(int i=l;i<l+(r-l)/2;i++) {
    		T temp = arr[i]; arr[i] = arr[r-i+l-1]; arr[r-i+l-1] = temp;
    	}
    }
    	
    	
    static class FastScanner{
    	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    	StringTokenizer st = new StringTokenizer("");
     
    	public String next(){
    		while(!st.hasMoreElements()){
    			try{
    				st = new StringTokenizer(br.readLine());
    			} catch(IOException e){
    				e.printStackTrace();
    			}
    		}
    		return st.nextToken();
    	}
    		
    	public String nextLine() throws IOException {
    		return br.readLine();
    	}
    		
    	public int nextInt(){
    		return Integer.parseInt(next());
    	}
     
    	public int[] readArray(int n){
    		int[] a = new int[n];
    		for(int i=0;i<n;i++)
    			a[i] = nextInt();
    		return a;
    	}
    		
    	public long nextLong() {
    		return Long.parseLong(next());
    	}
    		
    	public char nextChar() {
    		return next().toCharArray()[0];
    	}
    }


   	
}