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

class BIT:
  def __init__(self,n):
    self.n=n+1
    self.BIT=[0]*self.n
  def add(self,i,x):
    idx=i
    while idx<self.n:
      self.BIT[idx]+=x
      idx+=(idx&-idx)
  def _sum(self,i):
    ret=0
    idx=i
    while idx>0:
      ret+=self.BIT[idx]
      idx-=(idx&-idx)
    return ret
  def sum(self,l,r):
    return self._sum(r)-self._sum(l-1)
  def value(self,i):
    return self._sum(i)-self._sum(i-1)

n,q=map(int,input().split())
arr=list(map(int,input().split()))
query=[]
for i in range(q):
  l,r=map(int,input().split())
  query.append((l,r,i))
query=sorted(query,key=lambda x:x[0])
ans=[0]*q
bit=BIT(n)
pos_now=[-1]*(n+1)
pos_prev=[-1]*(n+1)
dic={}
for i in range(n):
  val=arr[i]
  dic[i+1]=-1
  if pos_now[val]==-1:
    bit.add(i+1,1)
    pos_now[val]=i+1
  if pos_prev[val]!=-1:
    dic[pos_prev[val]]=i+1
  pos_prev[val]=i+1
prev=1
for l,r,id in query:
  while prev<l:
    val=arr[prev-1]
    if dic[pos_now[val]]!=-1:
      pos_now[val]=dic[pos_now[val]]
      bit.add(pos_now[val],1)
    prev+=1
  ans[id]=bit.sum(l,r)
for val in ans:
  print(val)