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.BufferedInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.text.DecimalFormat;
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.InputMismatchException;
import java.util.LinkedList;
import java.util.Map.Entry;
import java.util.Queue;
import java.util.Random;
import java.util.Set;
import java.util.SortedSet;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;

public class Main
{	
	/********************************************** a list of common variables **********************************************/
	private MyScanner scan = new MyScanner();
	private PrintWriter out = new PrintWriter(System.out);
	private final double PI = Math.acos(-1.0);
	private final int SIZEN = (int)(1e5);
	private final int MOD = (int)(1e9 + 7);
	private final long MODH = 10000000007L, BASE = 10007;
	private final int[] DX = {0, 1, 0, -1}, DY = {-1, 0, 1, 0};
	private ArrayList<Integer>[] edge;
	private long[][] dp;
	
	public void build(int[] a, int left, int right, int level)
	{
		if(left + 1 == right)
		{
			return;
		}
		int mid = (left + right) >> 1;
		build(a, left, mid, level - 1);
		build(a, mid, right, level - 1);
		int[] tmp = new int[right - left];
		int i = left, j = mid, k = 0;
		while(i < mid && j < right)
		{
			if(a[i] < a[j])
			{
				dp[level][1] += right - j;
				tmp[k++] = a[i++];
			}
			else if(a[i] > a[j])
			{
				dp[level][0] += mid - i;
				tmp[k++] = a[j++];
			}
			else
			{
				int cnt1 = 0, cnt2 = 0, key = a[i];
				while(i < mid && a[i] == key)
				{
					++cnt1;
					++i;
				}
				while(j < right && a[j] == key)
				{
					++cnt2;
					++j;
				}
				for(int p = 0;p < cnt1 + cnt2;++p)
				{
					tmp[k++] = key;
				}
				dp[level][0] += (long)(mid - i) * cnt2;
				dp[level][1] += (long)(right - j) * cnt1;
			}
		}
		while(i < mid)
		{
			tmp[k++] = a[i++];
		}
		while(j < right)
		{
			tmp[k++] = a[j++];
		}
		System.arraycopy(tmp, 0, a, left, right - left);
	}

	public void foo()
	{
		int n = scan.nextInt();
		int total = 1 << n;
		int[] a = new int[total];
		for(int i = 0;i < total;++i)
		{
			a[i] = scan.nextInt();
		}
		dp = new long[n + 1][2];
		build(a, 0, total, n);
		int m = scan.nextInt();
		while(m-- > 0)
		{
			int q = scan.nextInt();
			for(int i = 0;i <= q;++i)
			{
				long tmp = dp[i][0];
				dp[i][0] = dp[i][1];
				dp[i][1] = tmp;
			}
			long ans = 0;
			for(int i = 0;i <= n;++i)
			{
				ans += dp[i][0];
			}
			out.println(ans);
		}
	}

	public static void main(String[] args)
	{
		Main m = new Main();
		m.foo();
		m.out.close();
	}
	
	/********************************************** a list of common algorithms **********************************************/
	/**
	 * 1---Get greatest common divisor
	 * @param a :	first number
	 * @param b :	second number
	 * @return		greatest common divisor
	 */
	public long gcd(long a, long b)
	{
		return 0 == b ? a : gcd(b, a % b);
	}
	
	/**
	 * 2---Get the distance from a point to a line
	 * @param x1	the x coordinate of one endpoint of the line
	 * @param y1	the y coordinate of one endpoint of the line
	 * @param x2	the x coordinate of the other endpoint of the line
	 * @param y2	the y coordinate of the other endpoint of the line
	 * @param x		the x coordinate of the point 
	 * @param y		the x coordinate of the point
	 * @return		the distance from a point to a line
	 */
	public double getDist(long x1, long y1, long x2, long y2, long x, long y)
	{
		long a = y2 - y1;
		long b = x1 - x2;
		long c = y1 * (x2 - x1) - x1 * (y2 - y1);
		return Math.abs(a * x + b * y + c) / Math.sqrt(a * a + b * b);
	}
	
	/**
	 * 3---Get the distance from one point to a segment (not a line)
	 * @param x1	the x coordinate of one endpoint of the segment
	 * @param y1	the y coordinate of one endpoint of the segment
	 * @param x2	the x coordinate of the other endpoint of the segment
	 * @param y2	the y coordinate of the other endpoint of the segment
	 * @param x		the x coordinate of the point
	 * @param y		the y coordinate of the point
	 * @return		the distance from one point to a segment (not a line)
	 */
	public double ptToSeg(long x1, long y1, long x2, long y2, long x, long y)
	{
		double cross = (x2 - x1) * (x - x1) + (y2 - y1) * (y - y1);
		if(cross <= 0)
		{
			return (x - x1) * (x - x1) + (y - y1) * (y - y1);
		}
		double d = (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1);
		if(cross >= d) 
		{
			return (x - x2) * (x - x2) + (y - y2) * (y - y2);
		}
		double r = cross / d;
		double px = x1 + (x2 - x1) * r;
		double py = y1 + (y2 - y1) * r;
		return (x - px) * (x - px) + (y - py) * (y - py);
	}
	
	/**
	 * 4---KMP match, i.e. kmpMatch("abcd", "bcd") = 1, kmpMatch("abcd", "bfcd") = -1.
	 * @param t:	String to match.
	 * @param p:	String to be matched.
	 * @return		if can match, first index; otherwise -1.
	 */
	public int kmpMatch(char[] t, char[] p)
	{
		int n = t.length;
		int m = p.length;
		int[] next = new int[m + 1];
		next[0] = -1;
		int j = -1;
		for(int i = 1;i < m;++i)
		{
			while(j >= 0 && p[i] != p[j + 1])
			{
				j = next[j];
			}
			if(p[i] == p[j + 1])
			{
				++j;
			}
			next[i] = j;
		}
		j = -1;
		for(int i = 0;i < n;++i)
		{
			while(j >= 0 && t[i] != p[j + 1])
			{
				j = next[j];
			}
			if(t[i] == p[j + 1])
			{
				++j;
			}
			if(j == m - 1)
			{
				return i - m + 1;
			}
		}
		return -1;
	}
	
	/**
	 * 5---Get the hash code of a String
	 * @param	s:	input string
	 * @return	hash code
	 */
	public long hash(String s)
	{
		long key = 0, t = 1;
		for(int i = 0;i < s.length();++i)
		{
			key = (key + s.charAt(i) * t) % MODH;
			t = t * BASE % MODH;
		}
		return key;
	}
	
	/**
	 * 6---Get x ^ n % MOD quickly.
	 * @param	x:	base
	 * @param 	n:	times
	 * @return	x ^ n % MOD
	 */
	public long quickMod(long x, long n)
	{
		long ans = 1;
		while(n > 0)
		{
			if(1 == n % 2)
			{
				ans = ans * x % MOD;
 			}
			x = x * x % MOD;
			n >>= 1;
		}
		return ans;
	}
	
	class MyScanner
	{
	    private byte[] buf = new byte[1024];
	    private int curChar;
	    private int numChars;
	    BufferedInputStream bis = new BufferedInputStream(System.in);

	    public int read() 
	    {
	        if (-1 == numChars)
	        {
	        	throw new InputMismatchException();
	        }
	        if (curChar >= numChars)
	        {
	            curChar = 0;
	            try 
	            {
	                numChars = bis.read(buf);
	            }
	            catch (IOException e)
	            {
	                throw new InputMismatchException();
	            }
	            if (numChars <= 0)
	            {
	            	return -1;
	            }
	        }
	        return buf[curChar++];
	    }

	    public int nextInt() 
	    {
	        int c = read();
	        while (isSpaceChar(c))
	        {
	        	c = read();
	        }
	        int sgn = 1;
	        if (c == '-') 
	        {
	            sgn = -1;
	            c = read();
	        }
	        int res = 0;
	        do 
	        {
	            if (c < '0' || c > '9')
	            {
	            	throw new InputMismatchException();
	            }
	            res *= 10;
	            res += c - '0';
	            c = read();
	        } while (!isSpaceChar(c));
	        return res * sgn;
	    }
	    
	    public long nextLong() 
	    {
	    	int c = read();
	    	while (isSpaceChar(c))
	    	{
	    		c = read();
	    	}
	    	int sgn = 1;
	    	if (c == '-') 
	    	{
	    		sgn = -1;
	    		c = read();
	    	}
	    	long res = 0;
	    	do 
	    	{
	    		if (c < '0' || c > '9')
	    		{
	    			throw new InputMismatchException();
	    		}
	    		res *= 10;
	    		res += c - '0';
	    		c = read();
	    	} while (!isSpaceChar(c));
	    	return res * sgn;
	    }
	    
	    public double nextDouble() 
	    {
	        int c = read();
	        while (isSpaceChar(c))
	        {
	        	c = read();
	        }
	        int sgn = 1;
	        if (c == '-')
	        {
	            sgn = -1;
	            c = read();
	        }
	        double res = 0;
	        while (!isSpaceChar(c) && c != '.') 
	        {
	            if (c == 'e' || c == 'E')
	            {
	            	return res * Math.pow(10, nextInt());
	            }
	            if (c < '0' || c > '9')
	            {
	            	throw new InputMismatchException();
	            }
	            res *= 10;
	            res += c & 15;
	            c = read();
	        }
	        if (c == '.') 
	        {
	            c = read();
	            double m = 1;
	            while (!isSpaceChar(c)) 
	            {
	                if (c == 'e' || c == 'E')
	                {
	                	return res * Math.pow(10, nextInt());
	                }
	                if (c < '0' || c > '9')
	                {
	                	throw new InputMismatchException();
	                }
	                m /= 10;
	                res += (c & 15) * m;
	                c = read();
	            }
	        }
	        return res * sgn;
	    }
	    
	    public String next()
	    {
	        int c = read();
	        while (isSpaceChar(c))
	        {
	        	c = read();
	        }
	        StringBuilder res = new StringBuilder();
	        do 
	        {
	            res.appendCodePoint(c);
	            c = read();
	        } while (!isSpaceChar(c));
	        return res.toString();
	    }

	    private boolean isSpaceChar(int c) 
	    {
	        return ' ' == c || '\n' == c || '\r' == c || '\t' == c || -1 == c;
	    }
	}
}