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 Ruby (2.7.1) (judge verdict: Time Limit Exceeded)

@n, q = gets.split.map(&:to_i)
c = gets.split.map(&:to_i)

m = Array.new(@n+1) {[]}
@n.times {|i| m[c[i]] << i + 1}

@f = Array.new(@n+1, 0)
def add(i, x)
    return if i == 0
    while i <= @n
        @f[i] += x
        i += i & -i
    end
end

def sum(i)
    s = 0
    while i > 0
        s += @f[i]
        i &= i - 1
    end
    s
end

m.each {|e| add(e[0], 1) if !e.empty?}

klr = q.times.map {|i| [i] + gets.split.map(&:to_i)}.sort_by!{|k,l,r| l}
ans = Array.new(q)
i = 0
klr.each do |k,l,r|
    while i + 1 < l
        add(m[c[i]].shift, -1)
        add(m[c[i]][0], 1) if !m[c[i]].empty?
        i += 1
    end

    ans[k] = sum(r) - sum(l-1)
end
puts ans