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 PyPy3 (7.3.0) (judge verdict: Runtime Error)

import sys
from bisect import bisect_right
from math import ceil
def input(): return sys.stdin.readline().strip()

class BIT:

    """
    https://tjkendev.github.io/procon-library/python/range_query/bit.html
    Binary index treeの実装
    配列[a1, a2,...,an]に対して以下のクエリをO(logn)で行う:
        1. aiにxを加える
        2. 区間和 ai + a(i+1) + ... + aj の和を求める
    isom法を使えば、
        1. aiの値を取得する
        2. 区間[i, j]の全ての数にxを加算する
    もO(logN)で行うことができる
    """

    def __init__(self, n):
        """
        添字は1スタート
        """
        self.n = n
        self.data = [0] * (n + 1)
        self.el = [0] * (n + 1)

    def add(self, i, x):
        """
        i>0に対してaiにxを加算(x < 0でもOK)
        """
        if i <= 0 or self.n < i:
            print("i should be within 1 to n")
            raise ValueError
        else:
            self.el[i] += x
            while i <= self.n:
                self.data[i] += x
                i += i & -i

    def sum(self, i):
        """
        添字1からiまでの累積和を求める
        """
        s = 0
        while i > 0:
            s += self.data[i]
            i -= i & -i # i $ (-i)でiの最下位ビットのみ立った値を得る
        return s

    def get(self, i, j=None):
        """
        添字iからjまでの累積和を求める
        j=Noneの場合はaiの値を返す
        """
        if j is None:
            return self.el[i]
        return self.sum(j) - self.sum(i - 1)


def main():
    """
    区間の種類数を答える問題は典型みたい。
    https://hama-du-competitive.hatenablog.com/entry/2016/10/01/001418
    ポイントはクエリをrの昇順に整理しておくこと。
    玉の各種類ごとに、それが最後にどこで出現したかをBITに記録していく。
    """
    N, Q = map(int, input().split())
    C = [-1] + list(map(int, input().split()))
    queri = []
    for q in range(Q):
        l, r = map(int, input().split())
        queri.append((r, l, q))
    queri.sort()

    lastAppeared = [-1] * N  # lastAppeared[x] : 値 x が最後に出現した位置
    r0, l0, _ = queri[0]
    for i, c in enumerate(C):
        if i == 0: continue
        if i > r0: break
        lastAppeared[c] = i
    tree = BIT(N)
    for x in range(N):
        if lastAppeared[x] != -1:
            tree.add(lastAppeared[x], 1)

    ans = [0] * Q
    pre_R = r0
    for r, l, q in queri:
        if r > pre_R:
            for i in range(pre_R + 1, r + 1):
                if lastAppeared[C[i]] != -1:
                    tree.add(lastAppeared[C[i]], -1)
                lastAppeared[C[i]] = i
                tree.add(i, 1)
        ans[q] = tree.get(l, r)

    for a in ans: print(a)

if __name__ == "__main__":
    main()