Problem: 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

Solution (in PYTHON):

Solution:
import sys

sys.setrecursionlimit(10 ** 6)

def pyes_no(condition) :
  if condition :
    print ("YES")
  else :
    print ("NO")

def plist(a, s = ' ') :
  print (s.join(map(str, a)))

def rint() :
  return int(sys.stdin.readline())

def rstr() :
  return sys.stdin.readline().strip()


def rints() :
  return map(int, sys.stdin.readline().split())

def rfield(n, m = None) :
  if m == None :
    m = n
  
  field = []
  for i in xrange(n) :
    chars = sys.stdin.readline().strip()
    assert(len(chars) == m)
    field.append(chars)
  return field

def pfield(field, separator = '') :
  print ('\n'.join(map(lambda x: separator.join(x), field)))

def check_field_equal(field, i, j, value) :
  if i >= 0 and i < len(field) and j >= 0 and j < len(field[i]) :
    return value == field[i][j]
  return None 

def digits(x, p) :
  digits = []
  while x > 0 :
    digits.append(x % p)
    x //= p
  return digits

def modpower(a, n, mod) :
  r = a ** (n % 2)
  if n > 1 :
    r *= modpower(a, n // 2, mod) ** 2
  return r % mod

def gcd(a, b) :
  if a > b :
    a, b = b, a
  
  while a > 0 :
    a, b = b % a, a

  return b

def vector_distance(a, b) :
  diff = vector_diff(a, b)
  
  return scalar_product(diff, diff) ** 0.5

def vector_inverse(v) :
  r = [-x for x in v]

  return tuple(r)

def vector_diff(a, b) :
  return vector_sum(a, vector_inverse(b))

def vector_sum(a, b) :
  r = [c1 + c2 for c1, c2 in zip(a, b)]
    
  return tuple(r)

def scalar_product(a, b) :
  r = 0
  for c1, c2 in zip(a, b) :
    r += c1 * c2

  return r

def check_rectangle(points) :
  assert(len(points) == 4)

  A, B, C, D = points

  for A1, A2, A3, A4 in [
    (A, B, C, D),
    (A, C, B, D),
    (A, B, D, C),
    (A, C, D, B),
    (A, D, B, C),
    (A, D, C, B),
  ] :
    sides = (
      vector_diff(A1, A2),
      vector_diff(A2, A3),
      vector_diff(A3, A4),
      vector_diff(A4, A1),
    )
    if all(scalar_product(s1, s2) == 0 for s1, s2 in zip(sides, sides[1:])) :
       return True
  return False

def check_square(points) :
  if not check_rectangle(points) :
    return False
  A, B, C, D = points

  for A1, A2, A3, A4 in [
    (A, B, C, D),
    (A, C, B, D),
    (A, B, D, C),
    (A, C, D, B),
    (A, D, B, C),
    (A, D, C, B),
  ] :
    side_lengths = [
      (first[0] - next[0]) ** 2 + (first[1] - next[1]) ** 2 for first, next in zip([A1, A2, A3, A4], [A2, A3, A4, A1])
    ]
    if len(set(side_lengths)) == 1 :
      return True
    
  return False

def check_right(p) :
  # Check if there are same points
  for a, b in [
    (p[0], p[1]),
    (p[0], p[2]),
    (p[1], p[2]),
  ] :
    if a[0] == b[0] and a[1] == b[1] :
      return False

  a, b, c = p
  a, b, c = vector_diff(a, b), vector_diff(b, c), vector_diff(c, a)   

  return scalar_product(a, b) * scalar_product(a, c) * scalar_product(b, c) == 0

n, m = rints()
a = rints()
counts = {}
values = {}
for i, c in enumerate(a[::-1]) :
  values[c] = True
  counts[n - i] = len(values)

for j in range(m) :
  index = rint()
  print counts[index]