Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

4 3
1 2 1 3
1 3
2 4
3 3

Sample Output 1

2
3
1

The 1-st, 2-nd, and 3-rd balls from the left have the colors 1, 2, and 1 - two different colors.

The 2-st, 3-rd, and 4-th balls from the left have the colors 2, 1, and 3 - three different colors.

The 3-rd ball from the left has the color 1 - just one color.

Sample Input 2

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

Sample Output 2

1
2
2
1
2
2
6
3
3
3

Solution in Python (3.8.2) (judge verdict: Time Limit Exceeded)

import numpy as np

class BIT:
	'''NOTES: 1-origin-indexed
	'''
	def __init__(self, itemCount):
		self.itemCount = itemCount
		self.items = np.zeros(itemCount+1,dtype=np.int)

	def add(self, i, value):
		while i <= self.itemCount:
			self.items[i] += value
			i += (i & (-i))

	def sumFromStart(self, end):
		summary = 0
		i = end
		while i > 0:
			summary += self.items[i]
			i -= (i & (-i))
		return summary

	def sum(self, start,end):
		summary = self.sumFromStart(end) - self.sumFromStart(start-1)
		return summary


N,Q = map(int,input().split())

C=list(map(int,input().split()))

# Query取得-rightでソート
dtypeQuery = [("index", int), ("start", int), ("end", int)]
queries=[]
for q in range(Q):
	l,r=map(int,input().split())
	queries.append( (q,l,r) )
queries=np.array(queries, dtype=dtypeQuery)
queries=np.sort(queries,order="end")

# 答え配列
ans=np.zeros(Q, dtype=np.int)
# 各色の現在の一番右のindex
mostRightColorIndex = np.zeros(N+1, dtype=np.int)
# bit indexed tree
bitree=BIT(N)
# 左からBITと現在色の更新をしながら、クエリのrightに到達したときにBITにクエリ発行する
qindex = 0
for n in range(N):
	# 外側のループでは、全部の色をループ

	if 0 < mostRightColorIndex[C[n]]:
		# ループ中の場所の色のindexが既にある場合(=過去に、同じ色が登録されていた場合)
		# 今回で、同じ色の一番右側の要素はループ中の要素になる
		# 前登録したこの色の一番右のindexを一度BITから抜く(↓で再度登録される)
		bitree.add(mostRightColorIndex[C[n]], -1 )

	# この要素の色の一番右のindexを、BITに登録
	mostRightColorIndex[C[n]] = n+1
	bitree.add(n+1, 1)

	while qindex < Q and n+1 == queries[qindex][2]:
		# 今のBITが次のクエリ発行するための状態(n==query.right)だったら、クエリ発行
		index = queries[qindex][0]
		start = queries[qindex][1]
		end = queries[qindex][2]
		ans[index]=bitree.sum(start,end)
		# print(index,start,end,ans[index])
		qindex += 1


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