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 Java (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
import java.util.HashSet;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Object o = run();
        if (o != null)
            System.out.println(o);
    }

    static Object run() {
        try (Scanner sc = new Scanner(System.in)) {
            int n = sc.nextInt(), q = sc.nextInt();
            ArrayList<Integer> c = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                c.add(sc.nextInt());
            }
            for (int i = 0; i < q; i++) {
                int l = sc.nextInt()-1, r = sc.nextInt()-1;
                HashSet<Integer> hs = new HashSet<>();
                for (int j = l; j <= r; j++) {
                    hs.add(c.get(j));
                }
                System.out.println(hs.size());
            }
            return null;
        }
    }
}

class ChenNumberUtil {
    static Collection<Integer> getDivisors(int n) {
        ArrayList<Integer> al = new ArrayList<>();
        for (int i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                al.add(i);
                if (n / i != i)
                    al.add(n / i);
            }
        }
        Collections.sort(al);
        return al;
    }
}

class ChenBitUtil {
    static int popCount(int bit) {
        return Integer.bitCount(bit);
    }

    static boolean isFlag(int bit, int i) {
        return 0 != (bit & (1 << i));
    }

    static int setTrue(int bit, int i) {
        return bit |= (1 << i);
    }

    static int setFalse(int bit, int i) {
        return bit &= ~(1 << i);
    }

    static int setReverse(int bit, int i) {
        return isFlag(bit, i) ? setFalse(bit, i) : setTrue(bit, i);
    }

    static int reverseAll(int bit) {
        return ~bit;
    }
}