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;
typedef long long ll;
#define N 500000
#define mod1 1000000007
#define mod2 1000000009
#define mod3 998244353
#define endl '\n'
#define IO ios_base::sync_with_stdio(false);cin.tie(NULL);
vector <ll> a(N+5),b(N+5);
vector <ll> segtree[4*N+5];
void build(ll ptr,ll start,ll end)
{
	if(start==end)
	{
		segtree[ptr].push_back(a[start]);
		return;
	}
	ll mid=(start+end)/2;
	build(2*ptr,start,mid);
	build(2*ptr+1,mid+1,end);
	ll pointer1=0,pointer2=0;
	while(pointer1<segtree[2*ptr].size() && pointer2<segtree[2*ptr+1].size())
	{
		if(segtree[2*ptr][pointer1]<segtree[2*ptr+1][pointer2])
		{
			segtree[ptr].push_back(segtree[2*ptr][pointer1]);
			pointer1++;
		}
		else
		{
			segtree[ptr].push_back(segtree[2*ptr+1][pointer2]);
			pointer2++;
		}
	}
	while(pointer1<segtree[2*ptr].size())
	{
		segtree[ptr].push_back(segtree[2*ptr][pointer1]);
		pointer1++;
	}
	while(pointer2<segtree[2*ptr+1].size())
	{
		segtree[ptr].push_back(segtree[2*ptr+1][pointer2]);
		pointer2++;
	}
	return;
}
ll query(ll ptr,ll l,ll r,ll val,ll start,ll end)
{
	if(start>r || l>end)
		return 0;
	else if(start>=l && r>=end)
	{
		ll count = lower_bound(segtree[ptr].begin(),segtree[ptr].end(),val)-segtree[ptr].begin();
		return (segtree[ptr].size()-count);
	}
	else
	{
		ll mid=(start+end)/2;
		return (query(2*ptr,l,r,val,start,mid) + query(2*ptr+1,l,r,val,mid+1,end));
	}
}
int main()
{
	IO
	clock_t begin = clock();
	ll n,q;
	cin>>n>>q;
	map <ll,ll> mp;
	for(ll i=1;i<=n;i++)
		cin>>a[i];
	for(ll i=1;i<=n;i++)
		b[i]=a[i];
	for(ll i=n;i>=1;i--)
	{
		a[i]=(mp[b[i]]==0?n+1:mp[b[i]]);
		mp[b[i]]=i;
	}
	// for(ll i=1;i<=n;i++)
	// 	cout<<a[i]<<' ';
	// cout<<endl;
	build(1,1,n);
	while(q--)
	{
		ll l,r;
		cin>>l>>r;
		ll ans=query(1,l,r,r+1,1,n);
		cout<<ans<<'\n';
	}
    // cout<<double(clock() - begin)/CLOCKS_PER_SEC<<endl;
    return 0;
}