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
## necessary imports
import sys
# import random
# from math import log2, log, ceil
input = sys.stdin.readline

# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp

## gcd function
def gcd(a,b):
    if a == 0:
        return b
    return gcd(b%a, a)

## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes

## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b

# find function
def find(x, link):
    while(x != link[x]):
        x = link[x]
    return x

# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, size, link):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x

## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    return prime

#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array

################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)] 
# spf_sieve() 
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

## taking integer array input
def int_array():
    return list(map(int, input().split()))

#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');


################# ---------------- TEMPLATE ENDS HERE ---------------- #################


## recursive knapsack function using memoization

n, m = int_array(); a = int_array(); ans = [];
this = set();
for i in range(n-1, -1, -1):
    if a[i] not in this:
        this.add(a[i]);
    ans.append(len(this));
ans = ans[::-1];
for _ in range(m):
    print(ans[int(input()) -1])