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

public class Main
{   
    static int[] countDistinctNumberFreq(int[] numArr)
    {                        
        int countDistinct = 0;
        HashMap<Integer,Integer> distinctNums = new HashMap<>();
        int[] cfreq = new int[numArr.length];
 
        for(int i=numArr.length-1; i>=0; i--)
        {
            if(distinctNums.get(numArr[i]) != null)
                distinctNums.put(numArr[i],distinctNums.get(numArr[i])+1);
	        else
	        {
	            distinctNums.put(numArr[i],1);
	            countDistinct++;
	        }
	        cfreq[i] = distinctNums.size();
        }
        return cfreq;
    }
    
    public static void main(String[] args) throws IOException 
	{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int m = Integer.parseInt(st.nextToken());
		
		int[] numArr = new int[n];
	 
		st = new StringTokenizer(br.readLine());
	    for(int i=0; i<n; i++)
	        numArr[i] = Integer.parseInt(st.nextToken());

    	int[] cmFreq = countDistinctNumberFreq(numArr);
    	for(int i=0; i<m; i++)
    	{
    	    st = new StringTokenizer(br.readLine());
    	    int lth = Integer.parseInt(st.nextToken());
    	    System.out.println(cmFreq[lth-1]);
    	}
    }

    static double findDistanceBetweenTwoPoints(int x1, int y1, int x2, int y2){
        return Math.sqrt(Math.pow(x2-x1,2) + Math.pow(y2-y1,2));
    }

    static int[] countLetterFreqeuncy(String str, int[] ltrs){
        for(int i=0; i<str.length(); i++)
            ltrs[(int) (str.charAt(i) - 'a')]++;
        return ltrs;
    }
	
    static long gcd(long a, long b) { 
        if (a == 0) 
            return b;  
        return gcd(b % a, a);  
    } 
      
    static long lcm(long a, long b) { 
        return (a*b)/
        gcd(a, b); 
    } 

    // Arrays.sort(arr, new Comparator<Integer[]>() {
    //     @Override
    //             //arguments to this method represent the arrays to be sorted   
    //     public int compare(Integer[] arr1, Integer[] arr2) {
    //             //get the item ids which are at index 0 of the array
    //             Integer item1 = arr1[0];
    //             Integer item2 = arr2[0];
    //         // sort based on price
    //         return item1.compareTo(item2);
    //     }
    // });

    //String.join("", Collections.nCopies(n-1, "");

    static int binarySearch(int left, int right, int find, int[] arr){
        int mid = left + (right-left)/2;
        if(find == arr[mid]) return mid;
        else if(find < arr[mid]) return binarySearch(left,mid-1,find,arr);
        else return binarySearch(mid+1,right,find,arr);
    }

    static ArrayList<Integer> separateDigits(int num){
         int pow = 0;
         ArrayList<Integer> arr = new ArrayList<>();
         while(num != 0)
         {
             if(num % 10 != 0)
             {
                 arr.add((num % 10) * (int) Math.pow(10, pow));
             }
             num /= 10;
             pow++;
         }
         return arr;
     }

     //find pattern
//     for(int i = 0; i<n-1; i++)
//     {
//         for(int j = 0; j<m-1; j++)
//         {
//                 if(pixel[i].charAt(j)*pixel[i+1].charAt(j)*pixel[i].charAt(j+1)*pixel[i+1].charAt(j+1) == 'f'*'a'*'c'*'e')
//                     faceNum++;
//         }
//     }


// row and column checking
    static boolean checkAllRows(String[] board)
    {
        for(int i=0; i<board.length; i++)
        {
            for(int j=0; j<board.length; j++)
            {
                if(board[i].charAt(j) == 'B')
                {
                    return false;
                } 
            }
        }
        return true;
    }

    

    class Pair implements Comparable<Pair>{
        int x;
        int y;
     
        public Pair(int x, int y)
        {
            this.x = x;
            this.y = y;
        }
        @Override
        public int compareTo(Pair p)
        {
            if(p.y == y)
            return x - p.x;
            return p.y - y;
        }
    }
}













