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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#pragma comment(linker, "/stack:200000000")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//turn on extra precision
//#pragma GCC target("fpmath=387")

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef string str;
typedef pair <int,int> pii;
typedef pair <ll,ll> pll;
typedef vector <int> vi;
typedef vector <ll> vll;
typedef vector <pii> vpii;
typedef vector <pll> vpll;

#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define ordered_multiset tree<int, null_type,less_equal<int>, rb_tree_tag,tree_order_statistics_node_update>
#define mp make_pair
#define pb push_back
#define pob pop_back
#define pf push_front
#define pof pop_front
#define fi first
#define se second
#define fs first.second
#define ss second.second
#define ff first.first
#define sf second.first
#define newl '\n'
#define fbo find_by_order
#define ook order_of_key
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(),x.rend()
#define watch(x) cout << (#x) << " is : " << (x) << newl

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
vi dirx = {0,0,1,-1};
vi diry = {1,-1,0,0};
char to_upper (char x){
    if( 97 <= int(x) && int(x) <= 122) return char(x-32);
    if( 65 <= int(x) && int(x) <= 90) return x;
    return -1;
}
char to_lower (char x){
    if( 97 <= int(x) && int(x) <= 122) return x;
    if( 65 <= int(x) && int(x) <= 90) return char(x+32);
    return -1;
}
int numerize (char x){
    if(48 <= int(x) && int(x) <= 57) return int(x-'0');
    if(97 <= int(x) && int(x) <= 122) return int(x-96);
    if(65 <= int(x) && int(x) <= 90) return int(x-64);
    return -1;
}
bool isect (int l1, int r1, int l2, int r2){
    pii p1 = {l1,r1};
    pii p2 = {l2,r2};
    if(p1>p2)swap(p1,p2);
    return (p2.fi <= p1.se);
}
ll quickpow (ll num1, ll num2, ll MOD){
    if(num2==0)return 1%MOD;
    else if(num2==1)return num1%MOD;
    else{
        ll temp = quickpow (num1,num2>>1LL,MOD); ll res = ((temp%MOD) * (temp%MOD))%MOD;
        if(num2&1) res = ((res%MOD)*(num1%MOD))%MOD; return res;
    }
}
ll invmod (ll num, ll MOD){return quickpow (num,MOD-2,MOD);}
ll gcd (ll num1, ll num2){
    if(num1 < num2) swap(num1,num2); ll num3 = num1 % num2 ;
    while(num3 > 0){ num1 = num2; num2 = num3; num3 = num1 % num2;}
    return num2;
}
ll lcm (ll num1 , ll num2){return (ll) (num1/__gcd(num1,num2))*num2;}
// end of Template
int BIT[500010],n,q,ans[500010],last[500010];
int getsum(int idx){
    int res = 0;
    for(int i = idx; i; i -= (i & -i)) res += BIT[i];
    return res;
}
void update(int idx, int val) {
    for(int i = idx; i <= n; i += (i & -i)) BIT[i] += val;
}
int main(){

//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    cin >> n >> q;
    vector < pair < pii , int > > tasks;
    for(int i = 1; i <= n; ++i) {
        int a; cin >> a;
        int occ = last[a];
        last[a] = i;
        tasks.pb({{occ,INT_MAX},i});
    }

    for(int i = 1; i <= q; ++i){
        int l, r; cin >> l >> r;
        tasks.pb({{l,r},i});
    }

    sort(all(tasks));

    for(auto t : tasks){
        if(t.fs == INT_MAX) {
            update(t.se,1);
            continue;
        }

        int idx = t.se;
        int l = t.ff;
        int r = t.fs;
        ans[idx] = getsum(r) - getsum(l-1);
    }

    for(int i = 1; i <= q; ++i) cout << ans[i] << newl;

    return 0;
}