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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::*;
use proconio::marker::*;

#[fastout]
fn run() {
    input! {
        n: usize,
        q: usize,
        c: [Usize1; n],
        ask: [(Usize1, usize); q],
    }
    let batch = 1024;
    let mut query = vec![vec![]; n / batch + 1];
    for (i, (l, r)) in ask.into_iter().enumerate() {
        query[l / batch].push((r, l, i));
    }
    let mut cnt = vec![0; n];
    let mut ans = vec![0; q];
    let mut sum = 0;
    let mut x = 0;
    let mut y = 0;
    for (i, mut query) in query.into_iter().enumerate() {
        query.sort_unstable_by_key(|p| p.0);
        if i & 1 == 1 {
            query.reverse();
        }
        for (r, l, k) in query {
            while l < x {
                x -= 1;
                let c = &mut cnt[c[x]];
                if *c == 0 {
                    sum += 1;
                }
                *c += 1;
            }
            while y < r {
                let c = &mut cnt[c[y]];
                y += 1;
                if *c == 0 {
                    sum += 1;
                }
                *c += 1;
            }
            while x < l {
                let c = &mut cnt[c[x]];
                x += 1;
                if *c == 1 {
                    sum -= 1;
                }
                *c -= 1;
            }
            while r < y {
                y -= 1;
                let c = &mut cnt[c[y]];
                if *c == 1 {
                    sum -= 1;
                }
                *c -= 1;
            }
            ans[k] = sum;
        }
    }
    for a in ans {
        println!("{}", a);
    }
}

fn main() {
    run();
}