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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define dbg puts("It's Just Begining.Wait For The End!!!!!\n")
#define CHECK(x) cout << (#x) << " is " << (x) <<"\n";
#define endl printf("\n")
typedef pair<ll,ll> pii;
#define pb push_back
#define ms(a,b) memset(a, b, sizeof(a))
#define in(x) scanf("%lld",&x)
#define out(x) printf("%lld",x)
#define sps printf(" ");
#define X first
#define Y second
#define dou(x) cout<<fixed<<setprecision(6)<<x
#define MAX 500005
ll K, ans[MAX], arr[MAX], sum,brr[MAX];
struct Query
{
    ll index, L, R;
    bool operator < (const Query &other) const
    {
        ll block_own = L/K;
        ll block_other = other.L/K;
        if(block_own == block_other)
            return R < other.R;
        return block_own < block_other;
    }
} query[MAX];
void add(ll idx)
{
    brr[arr[idx]]++;
    if(brr[arr[idx]]==1)sum++;
}
void remov(ll idx)
{
    brr[arr[idx]]--;
    if(brr[arr[idx]]==0)sum--;
}
int main()
{
    ll i,j,n,m,q;
    in(n);
    in(q);
    ms(brr,0);
    K=sqrt(n);
    for(i=1; i<=n; i++)in(arr[i]);
    for(i=0; i<q; i++)
    {
        in(query[i].L);
        in(query[i].R);
        query[i].index=i+1;
    }
    sort(query,query+q);
    ll l=1,r=0;
    sum=0;
    for(i=0; i<q; i++)
    {
        while(r<query[i].R)add(++r);
        while(l<query[i].L)remov(l++);
        while(r>query[i].R)remov(r--);
        while(l>query[i].L)add(--l);
        ans[query[i].index]=sum;
    }
    for(i=1; i<=q; i++)
    {
        out(ans[i]);
        endl;
    }
}