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 signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------


int N,Q;
int C[505050];
int L[505050],R[505050];
const int DI=800;
vector<pair<int,int>> ev[DI];
int ret[505050];
int num[505050];

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N>>Q;
	FOR(i,N) cin>>C[i];
	FOR(i,Q) {
		cin>>L[i]>>R[i];
		L[i]--;
		ev[L[i]/DI].push_back({R[i],i});
	}
	
	FOR(i,DI) {
		ZERO(num);
		int c=0;
		sort(ALL(ev[i]));
		int CL=0,CR=0;
		FORR(e,ev[i]) {
			x=e.second;
			while(CR<R[x]) {
				if(num[C[CR]]==0) c++;
				num[C[CR]]++;
				CR++;
			}
			while(CL<L[x]) {
				num[C[CL]]--;
				if(num[C[CL]]==0) c--;
				CL++;
			}
			while(L[x]<CL) {
				CL--;
				if(num[C[CL]]==0) c++;
				num[C[CL]]++;
			}
			ret[x]=c;
		}
		
	}
	FOR(i,Q) cout<<ret[i]<<endl;
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}