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
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from operator import itemgetter

N,Q=map(int,input().split())
C=[0]+list(map(int,input().split()))

QU=[list(map(int,input().split()))+[i] for i in range(Q)]


QU.sort(key=itemgetter(0))


# BIT(BIT-indexed tree)

LEN=N+3

BIT=[0]*(LEN+1) # 1-indexedなtree

def update(v,w): # index vにwを加える
    while v<=LEN:
        BIT[v]+=w
        v+=(v&(-v)) # 自分を含む大きなノードへ. たとえばv=3→v=4

def getvalue(v): # [1,v]の区間の和を求める
    ANS=0
    while v!=0:
        ANS+=BIT[v]
        v-=(v&(-v)) # 自分より小さい2ベキのノードへ. たとえばv=3→v=2へ
    return ANS

NUM=[[] for i in range(N+1)]
for i,c in enumerate(C):
    if i==0:
        continue
    if NUM[c]==[]:
        update(i,1)
    NUM[c].append(i)
    
NUMIND=[0]*(N+1)

ANS=[-1]*Q

ia=1

for x,y,ind in QU:

    while ia!=x:
        update(NUM[C[ia]][NUMIND[C[ia]]],-1)
        NUMIND[C[ia]]+=1
        if NUMIND[C[ia]]<len(NUM[C[ia]]):
            update(NUM[C[ia]][NUMIND[C[ia]]],1)
        ia+=1

    #print(x,y,ind)
    #print([getvalue(i) for i in range(1,N+1)])
    ANS[ind]=getvalue(y)

sys.stdout.write(" ".join(map(str,ANS))+"\n")