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 BinaryIndexedTree:
  def __init__(self,n):
    self.bit=[0]*n
  def add(self,i,x):
    i+=1
    while i<=len(self.bit):
      self.bit[i-1]+=x
      i+=i&-i
  def sum_sub(self,i):
    a=0
    i+=1
    while i:
      a+=self.bit[i-1]
      i-=i&-i
    return a
  def sum(self,i,j):
    a=self.sum_sub(j+1)
    if i!=0:a-=self.sum_sub(i-1)
    return a
n,q=map(int,input().split())
a=list(map(int,input().split()))
for i in range(n):a[i]-=1
queries=[[]for _ in range(n)]
for i in range(q):
  l,r=map(int,input().split())
  l-=1
  r-=1
  queries[r].append((l,i))
lastAppeared=[0]*n
Bit=BinaryIndexedTree(n+1)
ans=[-1]*q
for i in range(n):
  Bit.add(lastAppeared[a[i]],-1)
  Bit.add(i+1,1)
  lastAppeared[a[i]]=i+1
  for l,ind in queries[i]:
    ans[ind]=Bit.sum(l+1,i)
print(*ans)
