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: Time Limit Exceeded)

import io
import os


# From https://www.geeksforgeeks.org/mos-algorithm-query-square-root-decomposition-set-1-introduction/
def solve(N, Q, C, LR):
    order = sorted(range(len(LR)), key=lambda i: LR[i][1])

    currL, currR = 0, 0
    countUniq = [0 for i in range(N + 1)]
    numUniq = 0

    ans = [0] * len(LR)
    for i in order:
        L, R = LR[i]

        while currL < L:
            x = C[currL]
            countUniq[x] -= 1
            if countUniq[x] == 0:
                numUniq -= 1
            currL += 1

        while currL > L:
            x = C[currL - 1]
            if countUniq[x] == 0:
                numUniq += 1
            countUniq[x] += 1
            currL -= 1
        while currR <= R:
            x = C[currR]
            if countUniq[x] == 0:
                numUniq += 1
            countUniq[x] += 1
            currR += 1

        while currR > R + 1:
            x = C[currR - 1]
            countUniq[x] -= 1
            if countUniq[x] == 0:
                numUniq -= 1
            currR -= 1

        # print(C[L : R + 1], "is", [i for i, x in enumerate(countUniq) if x > 0], numUniq)
        ans[i] = numUniq
    print("\n".join(map(str, ans)))


if False:
    import random

    random.seed(0)
    N = 5 * 10 ** 5
    Q = 5 * 10 ** 5
    C = [random.randint(1, N) for i in range(N)]
    LR = []
    for i in range(Q):
        l = random.randint(1, N)
        r = random.randint(1, N)
        if l > r:
            l, r = r, l
        LR.append((l, r))
    solve(N, Q, C, LR)
    exit()


if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    N, Q = [int(x) for x in input().split()]
    C = [int(x) for x in input().split()]
    LR = [[int(x) - 1 for x in input().split()] for i in range(Q)]  # 0-indexed
    solve(N, Q, C, LR)