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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std ;

const int MAX = 5e5 + 10 ;

int arr[MAX] , last[MAX] , prv[MAX] ;
vector<int>tree[4 * MAX] ;

int n , q ;

void build(int node , int l , int r)
{
	if(l == r)
	{
		tree[node].push_back(prv[l]) ;
		return ;
	}
	int mid = (l + r) >> 1 ;
	build(node << 1 , l , mid) ;
	build(node << 1 | 1 , mid+1 , r) ;
	merge(tree[node << 1].begin() , tree[node << 1].end() , tree[node << 1 | 1].begin() , tree[node << 1 | 1] .end() , back_inserter(tree[node])) ;
}

int query(int node , int l , int r , int from , int to)
{
	if(from > r || to < l)
		return 0 ;
	if(l >= from && r <= to)
	{
		int x = lower_bound(tree[node].begin() , tree[node].end() , from) - tree[node].begin() ;
		return x ;
	}
	int mid = (l + r) >> 1 ;
	int a = query(node << 1 , l , mid , from , to) ;
	int b = query(node << 1 | 1 , mid+1 , r , from , to) ;
	return a+b ;
}

int main()
{
	ios_base::sync_with_stdio(0) ;
	cin.tie(0) ;
	cin>>n>>q ;
	for(int i = 1 ; i <= n ; ++i)
	{
		cin>>arr[i] ;
		prv[i] = last[arr[i]] ;
		last[arr[i]] = i ;
	}
	build(1 , 1 , n) ;
	while(q--)
	{
		int l , r ;
		cin>>l>>r ;
		cout<<query(1 , 1 , n , l , r)<<"\n" ;
	}
	return 0 ;
}