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
def main():
  def add(i,x):
    i+=1
    while i<=n:
      bit[i-1]+=x
      i+=i&-i
  def sum_sub(i):
    a=0
    i+=1
    while i:
      a+=bit[i-1]
      i-=i&-i
    return a
  def sum(i,j):
    a=sum_sub(j+1)
    if i!=0:a-=sum_sub(i-1)
    return a
  
  n,q,*x=map(int,open(0).read().split())
  a=x[:n]
  queries=[]
  for i in range(q):
    l,r=x[n+i*2:n+(i+1)*2]
    l-=1
    r-=1
    queries.append(r<<40|l<<20|i)
  ans=[0]*q
  queries.sort()
  for i in range(q):
    ind=queries[i]%(1<<20)
    queries[i]>>=20
    l=queries[i]%(1<<20)
    queries[i]>>=20
    r=queries[i]
    queries[i]=[r,l,ind]
  queries.append([10*20])
  lastAppeared=[-1]*n
  bit=[0]*n
  j=0
  for i in range(n):
    if lastAppeared[a[i]-1]!=-1:
      add(lastAppeared[a[i]-1],-1)
      add(i,1)
      lastAppeared[a[i]-1]=i
    else:
      add(i,1)
      lastAppeared[a[i]-1]=i
    while queries[j][0]==i:
      r,l,ind=queries[j]
      j+=1
      ans[ind]=sum(l,r-1)
  print(*ans)
if __name__ == "__main__":
  main()