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;
using ll = long long;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int N, Q;
    cin >> N >> Q;
    vector<int> C(N);
    for(int i=0; i<N; i++) cin >> C[i];
    vector<int> L(Q), R(Q);
    for(int i=0; i<Q; i++)
    {
        cin >> L[i] >> R[i];
        L[i]--; R[i]--;
    }

    vector<pair<set<int>, map<int, int>>> mp;
    set<int> s;
    map<int, int> m;
    for(int i=0; i<N; i++)
    {
        s.insert(C[i]);
        m[C[i]] = i;
        mp.emplace_back(s, m);
    }

    for(int i=0; i<Q; i++)
    {
        if(L[i] == 0)
        {
            cout << mp[R[i]].first.size() << '\n';
            continue;
        }
        set<int>& ls = mp[L[i]-1].first;
        set<int>& rs = mp[R[i]].first;
        map<int, int>& lm = mp[L[i]-1].second;
        map<int, int>& rm = mp[R[i]].second;
        int ans = rs.size();
        for(auto itr = ls.begin(); itr != ls.end(); itr++)
        {
            if(lm[*itr] == rm[*itr]) ans--;
        }
        cout << ans << '\n';
    }

    return 0;
}