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 = [int(i) for i in input().split()]
lr = [[int(i) - 1 for i in input().split()] for _ in range(Q)]

class BinaryIndexedTree() :
  def __init__(self, n) :
    self.n = n
    self.tree = [0] * (n + 1)
      
  def sum(self, i) :
    s = 0
    while i > 0 :
      s += self.tree[i]
      i -= i & -i
    return s
      
  def add(self, i, x) :
    while i <= self.n :
      self.tree[i] += x
      i += i & -i

BIT = BinaryIndexedTree(N + 5)
sch = [[i, l, r] for i, (l, r) in enumerate(lr)]
sch.sort(key=lambda x: x[2])
ret = [0] * Q
last = [-1] * (N + 1)

cur = 0
for i, l, r in sch :
  while cur <= r :
    BIT.add(cur + 1, 1)
    pos = last[c[cur]]
    if pos != -1 :
      BIT.add(pos + 1, -1)
    last[c[cur]] = cur
    cur += 1
  ret[i] = BIT.sum(r + 1) - BIT.sum(l)
for r in ret :
  print(r)