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: Accepted)

class BinaryIndexedTree:
    def __init__(self, size=10**7):
        self.size = size
        self.data = [0]*(size+1)

    def add(self, index, x):
        while index <= self.size:
            self.data[index] += x
            index += index & -index

    def prefix_sum(self, index):
        res = 0
        while index >= 1:
            res += self.data[index]
            index -= index & -index
        return res
    
    def interval_sum(self, left, right):
        return self.prefix_sum(right) - self.prefix_sum(left-1)

    def __getitem__(self, index):
        return self.sum(index) - self.sum(index-1)

    # def __setitem__(self, index, x):
    #     before = self.__getitem__(index)
    #     self.add(index, -before)
    #     self.add(index, x)

def main():
    from sys import stdin
    input = lambda: stdin.readline()
    LMIIS = lambda: list(map(int,input().split()))


    N,Q = LMIIS()
    # 1-indexedにするためリストの先頭に0を追加
    C = [0] + LMIIS()
    queries = []

    # 順番と一緒にクエリを保存
    for i in range(Q):
        l,r = map(int,input().split())
        queries.append((i,l,r))
    
    # rについてクエリをソート
    queries.sort(key=lambda x: x[2])

    # 玉の種類 -> 良い玉の場所
    good_ball_places = [0] * (N+1)

    # 場所 -> 良い玉が存在するなら1,しないなら0
    good_ball_exists = BinaryIndexedTree(N)

    #　玉の更新済み範囲
    k = 0
    # クエリの回答
    out = [0] * Q
    for order,l,r in queries:
        for i in range(k+1,r+1):
            # 位置iの玉の色
            color = C[i]
            # 色colorの良い玉の前の位置
            before = good_ball_places[color]
            # 色colorが既出なら、前の位置での存在を消去
            if 0 < before:
                good_ball_exists.add(before,-1)
            # 色colorの良い玉の位置はi
            good_ball_places[color] = i
            # 位置iに良い玉が存在
            good_ball_exists.add(i,1)
        # 良い玉を更新した範囲
        k = r
        # [l,r]に存在する良い玉の数の合計
        out[order] = good_ball_exists.interval_sum(l,r)
    print('\n'.join(map(str,out)))


        








    
main()