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: Wrong Answer)

#include <iostream>
#include<bits/stdc++.h>
using namespace std;

int main()
{
   long long int n,q ;
   cin>>n>>q ;
   vector<long long int > arr(n) ;
    for(int i=0 ; i<n ; i++)
        cin>>arr[i] ;
    unordered_map<long long int ,long long int> mp ;
    vector<long long> prefix(n,0) ;
    vector<long long> prefix_num(n,0) ;
    prefix[0]=0 ;
    mp[arr[0]]=0 ;
    prefix_num[0]=-1 ;
    for(int i=1 ; i<n ;i++)
    {
        if(mp.find(arr[i])!=mp.end())
        {
            prefix[i]=1+prefix[i-1] ;
            long long p=mp[arr[i]] ;
            prefix_num[i]=max(prefix_num[i-1],p) ;
        }
        else
        {
            prefix[i]=prefix[i-1] ;
            prefix_num[i]=prefix_num[i-1] ;
        }
        mp[arr[i]]=i ;
    }
    
    for(int i=1;i<=q;i++)
    {
        long long x,y ;
        cin>>x>>y ;
        x=x-1 ;
        y=y-1 ;
        if(x==0)
            cout<<(y-x+1-prefix[y])<<endl ;
        else if(x==y)
            cout<<1<<endl ;
        else
        {
            long long temp=prefix[y]-prefix[x-1] ;
            if(prefix_num[y]<x )
              cout<<(y-x+1)<<endl ;
            else if(prefix[y]!=prefix[x-1])
                cout<<(y-x+1 - temp )<<endl ;
            else
                cout<<(y-x+1)<<endl ;
        }
    }
    return 0;
}