Sereja has an array a, consisting of n integers a1, a2, ..., an. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out m integers l1, l2, ..., lm (1 ≤ li ≤ n). For each number li he wants to know how many distinct numbers are staying on the positions li, li + 1, ..., n. Formally, he want to find the number of distinct numbers among ali, ali + 1, ..., an.?

Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each li.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the array elements.

Next m lines contain integers l1, l2, ..., lm. The i-th line contains integer li (1 ≤ li ≤ n).

Output

Print m lines — on the i-th line print the answer to the number li.

Examples

Input

10 10
1 2 3 4 1 2 3 4 100000 99999
1
2
3
4
5
6
7
8
9
10


Output

6
6
6
6
6
5
4
3
2
1
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
    public static class FastIO
	{
		BufferedReader br;
		BufferedWriter bw, be;
		StringTokenizer st;

		public FastIO()
		{
			br = new BufferedReader(new InputStreamReader(System.in));
			bw = new BufferedWriter(new OutputStreamWriter(System.out));
			be = new BufferedWriter(new OutputStreamWriter(System.err));
			st = new StringTokenizer("");
		}

		private void read() throws IOException
		{
			st = new StringTokenizer(br.readLine());
		}

		public String ns() throws IOException
		{
			while(!st.hasMoreTokens())
				read();
			return st.nextToken();
		}

		public int ni() throws IOException
		{
			return Integer.parseInt(ns());
		}

		public long nl() throws IOException
		{
			return Long.parseLong(ns());
		}

		public float nf() throws IOException
		{
			return Float.parseFloat(ns());
		}

		public double nd() throws IOException
		{
			return Double.parseDouble(ns());
		}

		public char nc() throws IOException
		{
			return ns().charAt(0);
		}

		public int[] nia(int n) throws IOException
		{
			int[] a = new int[n];
			for(int i = 0; i < n; i++)
				a[i] = ni();

			return a;
		}

		public long[] nla(int n) throws IOException
		{
			long[] a = new long[n];
			for(int i = 0; i < n; i++)
				a[i] = nl();

			return a;
		}

		public void out(String s) throws IOException
		{
			bw.write(s);
		}

		public void flush() throws IOException
		{
			bw.flush();
			be.flush();
		}

		public void err(String s) throws IOException
		{
			be.write(s);
		}
	}
	 static boolean isPrime(int n)
    {
     // Check if number is less than
       // equal to 1
        if (n <= 1)
            return false;
        // Check if number is 2
        else if (n == 2)
            return true;
        // Check if n is a multiple of 2
        else if (n % 2 == 0)
            return false;
        // If not, then just check the odds
        for (int i = 3; i <= Math.sqrt(n); i += 2) 
        {
            if (n % i == 0)
                return false;
        }
        return true;
    }
     static int gcd(int a, int b)
    {
        if (a == 0)
            return b; 
        return gcd(b % a, a); 
    }
     
    // method to return LCM of two numbers
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
    
    static int smallestDivisor(int n)  
{  
    // if divisible by 2  
    if (n % 2 == 0)  
        return 2;  
  
    // iterate from 3 to sqrt(n)  
    for (int i = 3; i * i <= n; i += 2) {  
        if (n % i == 0)  
            return i;  
    }  
  
    return n;  
}  

	static FastIO f;
	
	public static void main(String args[]) throws IOException
	{ int i;
		f = new FastIO();
int a[]=new int[100001];
		int n,m;n=f.ni();m=f.ni();
		int res[]=new int [n+1];int ar[]=new int[n+1];
		for(i=1;i<=n;i++){
		    ar[i]=f.ni();
		    a[ar[i]]=0;
		    
		}
		res[n]=1;a[ar[n]]=1;
		for(i=n-1;i>0;i--){
		    if(a[ar[i]]==0){res[i]=res[i+1]+1; a[ar[i]]=1;}
		    else res[i]=res[i+1];
		}
		int temp;
		for(i=0;i<m;i++){ temp=f.ni();
   f.out(res[temp]+"\n");}
		f.flush();
	}
}
