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)

#pragma GCC target ("avx2")
#pragma GCC optimize ("unroll-loops")
#pragma GCC optimize ("O3")
#include "bits/stdc++.h"
#include <unordered_set>
#include <unordered_map>
#include <random>
using namespace std;
typedef long long ll;
const ll MOD = 1'000'000'007LL; /*998'244'353LL;*/
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rep(i, n) for(int (i)=0; (i)<(n); (i)++)
const int dx[4]={ 1,0,-1,0 };
const int dy[4]={ 0,1,0,-1 };

int N, Q;
int C[500000];
vector<pair<int,pair<int, int>>> query[800];
int ans[500000] ={};

signed main(){
    cin >> N >> Q;
    rep(i, N){
        cin >> C[i];
        C[i]--;
    }
    rep(i, Q){
        int l, r;
        cin >> l >> r;
        query[(l-1)/800].pb({ r, {l-1,i} });
    }

    rep(i, 800){
        vector<int> cnt(N, 0);
        int tmp = 0;
        int L = 800*i;
        int R = 800*i;
        sort(all(query[i]));
        rep(j, query[i].size()){
            int l = query[i][j].second.first;
            int r = query[i][j].first;
            int a = query[i][j].second.second;
            while(R < r){
                cnt[C[R]]++;
                if(cnt[C[R]] == 1) tmp++;
                R++;
            }
            while(L < l){
                cnt[C[L]]--;
                if(cnt[C[L]] == 0) tmp--;
                L++;
            }
            while(L > l){
                cnt[C[L-1]]++;
                if(cnt[C[L-1]] == 1) tmp++;
                L--;
            }
            ans[a] = tmp;
        }
    }

    rep(i, Q) cout << ans[i] << endl;
}