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
class Bit:
  def __init__(self,n):
    self.bl = n.bit_length()
    self.size = 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.size:
      self.tree[i] += x
      i += i & -i

import sys
input = sys.stdin.readline
from collections import defaultdict
n,q = map(int,input().split())
a = list(map(int,input().split()))
lr = [tuple(map(int,input().split()))+(i,) for i in range(q)]
ans = [0]*q
lr.sort(key = lambda x:x[1])
bit = Bit(n+2)
dc = defaultdict(int)
cnt = 0
for L,R,p in lr:
  L -= 1
  for i in range(cnt,R):
    if dc[a[i]]:
      l = dc[a[i]]-1
      r = i
      bit.add(l+2,1)
      bit.add(r+2,-1)
    else:
      bit.add(1,1)
      bit.add(i+2,-1)
    dc[a[i]] = i+1
  ans[p] = bit.sum(L+1)
  cnt = R
print(*ans,sep="\n")