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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
template <typename T>
struct BIT{
    int n;//元の数列のサイズ
    vector<T> bit[2];//データ収納用
    void init(int n_){
        n=n_+1;
        for(int j=0;j<2;j++){
            bit[j].assign(n,0);
        }
    }
    void add_sub(int p,int i,T x){
        for(int j=i;j<n;j+=j&-j){
            bit[p][j]+=x;
        }
    }
    void add(int l,int r,T x){//[l,r]に加算
        add_sub(0,l,-x*(l-1));
        add_sub(0,r+1,x*r);
        add_sub(1,l,x);
        add_sub(1,r+1,-x);
    }
    T sum_sub(int p,int i){
        T s(0);
        for (int j=i;j>0;j-=j&-j){
            s+=bit[p][j];
        }
        return s;
    }
    T sum(int i,int j){//[i,j]のΣを計算
        return sum_sub(0,j)+sum_sub(1,j)*j-sum_sub(0,i-1)-sum_sub(1,i-1)*(i-1);
    }
};
int main(){
    int N, Q;
    cin >> N >> Q;
    vector<int> C(N);
    for (int i = 0; i < N; i++) {
        cin >> C.at(i);
    }
    map<pair<int, int>, int> query;
    vector<vector<int>> left(N);
    for (int i = 0; i < Q; i++) {
        int l, r;
        cin >> l >> r;
        query[make_pair(l - 1, r - 1)] = i;
        left.at(r - 1).push_back(l - 1);
    }
    BIT<int> rock;
    rock.init(N);
    /*
    vector<int> RP(N, -1); //i番目の色の右端はどこか,初期値-1は存在しないことを示す
    vector<int> answer(N);
    for (int i = 0; i < N; i++) {
        int memo = RP.at(C.at(i) - 1);
        RP.at(C.at(i) - 1) = i;
        if (memo == -1) {
            rock.add(i + 1, i + 1, 1);
        }
        else {
            rock.add(memo + 1, memo + 1, -1);
            rock.add(i + 1, i + 1, 1);
        }
        for (int j = 0; j < (int) left.at(i).size(); j++) {
            int L = left.at(i).at(j);
            int number = query[make_pair(L, i)];
            answer.at(number) = rock.sum(L + 1, i + 1);
        }
    }
    for (int i = 0; i < Q; i++) {
        cout << answer.at(i) << endl;
    }
    */
}