We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.

You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?

Constraints

* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print Q lines. The i-th line should contain the response to the i-th query.

Examples

Input

4 3
1 2 1 3
1 3
2 4
3 3


Output

2
3
1


Input

10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8


Output

1
2
2
1
2
2
6
3
3
3
import sys
input = sys.stdin.readline

N, Q = map(int, input().split())
C = list(map(int, input().split()))
query = []

for i in range(Q):
  l, r = map(int, input().split())
  query.append((i, l, r))

query.sort(key = lambda x: x[2])

class BIT:
  from operator import add, sub
  X_unit = 0
  X_f = add
  X_f_rev = sub

  def __init__(self, seq):
    N = len(seq)
    self.N = N
    self.X = seq[:]
    for i in range(self.N):
      j = i + ((i+1) & -(i+1))
      if j < self.N:
        self.X[j] = self.X_f(self.X[i], self.X[j])
  
  def set_val(self, i, x):
    while(i < self.N):
      self.X[i] = self.X_f(self.X[i], x)
      i += (i+1) & -(i+1)

  def cum_val(self, i):
    res = self.X_unit
    while(i > -1):
      res = self.X_f(res, self.X[i])
      i -= (i+1) & -(i+1)
    return res

  def fold(self, L, R):
    return self.X_f_rev(self.cum_val(R-1), self.cum_val(L-1))

GJ = [-1]*N
bit = BIT([0]*N)
ans = [0]*Q
st = 0

for i in range(N):
  if GJ[C[i]-1] != -1:
    bit.set_val(GJ[C[i]-1], -1)
  GJ[C[i]-1] = i
  bit.set_val(i, 1)
  if st < Q:
    while query[st][2] == i+1:
      ans[query[st][0]] = bit.fold(query[st][1]-1, query[st][2])
      st += 1
      if st == Q:
        break

print(*ans, sep="\n")