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 = tuple(map(int,input().split()))
qs = [[] for _ in range(n)]
for i in range(q):
  l,r = map(int,input().split())
  l -= 1
  r -= 1
  qs[l].append((r,i))

class BIT:
    def __init__(self,n):
        self.num=n
        self.dat=[0]*(self.num+1)
        self.depth=n.bit_length()
    
    def add(self,i,x):
        i+=1
        while i<=self.num:
            self.dat[i]+=x
            i+=i&-i
    
    def sum(self,i):
        i+=1
        s=0
        while i>0:
            s+=self.dat[i]
            i-=i&-i
        return s
    
    def lower_bound(self,x):
        sum_=0
        pos=0
        for i in range(self.depth,-1,-1):
            k=pos+(1<<i)
            if k<=self.num and sum_+self.dat[k]<x:
                sum_+=self.dat[k]
                pos+=1<<i
        return pos, sum_

ps = [-1 for _ in range(n)]
pi = [-1 for _ in range(n+1)]
for i in range(n):
  l = pi[c[i]]
  if l != -1:
    ps[l] = i
  pi[c[i]] = i
    
d = BIT(n)
ans = [0 for _ in range(q)]
for x in range(n-1,-1,-1):
  if ps[x] != -1:
    d.add(ps[x],1)
  for r,i in qs[x]:
    ans[i] = (r-x+1)-d.sum(r)

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