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;
#define int long long int
#define mp(a,b) make_pair(a,b)
#define vi vector<int>
#define mii map<int,int>
#define mpi map<pair<int,int>,int>
#define vp vector<pair<int,int> >
#define pb(a) push_back(a)
#define fr(i,n) for(i=0;i<n;i++)
#define rep(i,a,n) for(i=a;i<n;i++)
#define F first
#define S second
#define endl "\n"
#define fast std::ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define mod 1000000007
#define dom 998244353
#define pi 3.14159265
#define sl(a) (int)a.length()
#define sz(a) (int)a.size()
#define all(a) a.begin(),a.end()
#define pii pair<int,int> 
#define mini 2000000000000000000
#define time_taken 1.0 * clock() / CLOCKS_PER_SEC
//mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
struct Query 
{ 
    int l, r, idx; 
};
bool cmp(Query x, Query y) 
{ 
    return x.r < y.r; 
} 
void update(int idx, int val, int bit[], int n) 
{ 
    for (; idx <= n; idx += idx&-idx) 
        bit[idx] += val; 
} 
int query(int idx, int bit[], int n) 
{ 
    int sum = 0; 
    for (; idx>0; idx-=idx&-idx) 
        sum += bit[idx]; 
    return sum; 
} 
void shikhar7s(int cas)
{
    int n,i,q;
    cin>>n>>q;
    int arr[n];
    fr(i,n)
    {
        cin>>arr[i];
    }
    Query queries[q];
    int x,y;
    fr(i,q)
    {
        cin>>x>>y;
        x--;
        y--;
        queries[i].l=x;
        queries[i].r=y;
        queries[i].idx=i;
    }
    sort(queries, queries+q, cmp); 
    int bit[n+1]; 
    memset(bit, 0, sizeof(bit)); 
    int last_visit[500005]; 
    memset(last_visit, -1, sizeof(last_visit)); 
    int ans[q]; 
    int query_counter = 0; 
    for (int i=0; i<n; i++) 
    { 
        if (last_visit[arr[i]] !=-1) 
            update (last_visit[arr[i]] + 1, -1, bit, n); 
        last_visit[arr[i]] = i; 
        update(i + 1, 1, bit, n); 
        while (query_counter < q && queries[query_counter].r == i) 
        { 
            ans[queries[query_counter].idx] = query(queries[query_counter].r + 1, bit, n)- query(queries[query_counter].l, bit, n); 
            query_counter++; 
        } 
    } 
    for (int i=0; i<q; i++) 
        cout << ans[i] << endl; 
}
signed main()
{
    fast;
    int t=1;
    //cin>>t;
    int cas=1;
    while(cas<=t)
    {
    shikhar7s(cas);
    cas++;
    }
    return 0;
}