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
n,q=map(int,input().split())
c=list(map(int,input().split()))
l=[list(map(int,input().split())) for i in range(q)]
for i in range(q):
  l[i].append(i)
l.sort(key=lambda x:x[1])

L=[-1]*(5*10**5+1)

class Bit:
  def __init__(self,n):
    self.size=n
    self.tree=[0]*(n + 1)
    self.depth=n.bit_length()

  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

BIT=Bit(n+1)

ans=[0]*q
ct=0
for i in range(q):
  while ct<=l[i][1]:
    if L[c[ct-1]-1]!=-1:
      BIT.add(L[c[ct-1]-1],-1)
    L[c[ct-1]-1]=ct+1
    BIT.add(ct+1,1)
    ct+=1
  ans[l[i][2]]=BIT.sum(l[i][1]+1)-BIT.sum(l[i][0])

for i in range(q):
  print(ans[i])