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;
typedef long long ll;
template<typename T1,typename T2> bool chmin(T1 &a,T2 b){if(a<=b)return 0; a=b; return 1;}
template<typename T1,typename T2> bool chmax(T1 &a,T2 b){if(a>=b)return 0; a=b; return 1;}
int dx[4]={0,1,-1,0};
int dy[4]={1,0,0,-1};
#define foR for

// FHCまでにスタックオーバーフローの罠を解決する

template<typename T>
struct BIT{
    vector<T> bit;
    BIT(int n) : bit(n+1) {}
    void add(int a,T w){
        for(int x=++a; x < bit.size(); x+=(x&(-x)) ){
            bit[x] += w;
        }
    }
    // return v[0]+v[1]+...+v[a-1];
    T sum(int a){
        T ret=0;
        for(int x=a;x>0;x-=(x&(-x)) ){
            ret+=bit[x];
        }
        return ret;
    }
};

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(20);

    int n,q;
    cin>>n>>q;
    BIT<int> b(n+1);
    int c[n];
    for(int i=0;i<n;i++){
        cin>>c[i];
        c[i]--;
    }
    array<int,3> a[q];
    int ans[q]={};
    for(int i=0;i<q;i++){
        int l,r;
        cin>>l>>r;
        a[i] = {r,l,i};
    }
    int pre[n]={};
    sort(a,a+q);
    int now = 0;
    for(int i=0;i<n;i++){
        if(pre[c[i]]){
            b.add(pre[c[i]],-1);
        }
        b.add(i+1,1);
        pre[c[i]] = i+1;
        while(now<q && a[now][0]==i+1){
            ans[a[now][2]] = b.sum(a[now][0]+1) - b.sum(a[now][1]);
            now++;
        }
    }
    for(int i=0;i<q;i++)cout << ans[i] << "\n";
}