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

def BIT_query(BIT,idx):
  res_sum=0
  if idx==0:
    return 0
  while idx>0:
    res_sum+=BIT[idx]
    idx-=idx&(-idx)
  return res_sum

def BIT_update(BIT,idx,x,n):
  while idx<=n:
    BIT[idx]+=x
    idx+=idx&(-idx)
  return

def main():
  N,Q=map( int, input().split())
  C=list( map( int, input().split()))
  LRI=[]
  for i in range(Q):
    l,r=map( int, input().split())
    LRI.append((r,l,i))
  LRI.sort(key=lambda x:x[0])
  lastAppend=[-1]*(N+1)
  BIT=[0]*(N+1)
  ANS=[0]*Q
  now=1
  for r,l,i in LRI:
    while now <= r:
      c=C[now-1]
      if lastAppend[c]== -1:
        BIT_update(BIT,now,1,N)
      else:
        BIT_update(BIT,now,1,N)
        BIT_update(BIT,lastAppend[c],-1,N)
      lastAppend[c]=now                
      now+=1
    ANS[i] = BIT_query(BIT,r) - BIT_query(BIT,l-1)
  for ans in ANS:
    print(ans)
    
if __name__ == '__main__':
    main()