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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <map>
#include <numeric>
#include <cmath>
#include <set>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <complex>
#include <string.h>
#include <unordered_set>
#include <unordered_map>
#include <bitset>
#include <iomanip>
#include <sys/time.h>
#include <tuple>
#include <random>
using namespace std;

#define endl '\n'
#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
#define UNIQ(v) (v).erase(unique((v).begin(), (v).end()), (v).end())

typedef long long ll;
typedef long double ld;
typedef pair<int, int> P;
typedef tuple<int, int, int> T;
typedef complex<double> comp;
typedef vector< vector<ld> > matrix;
struct pairhash {
public:
    template<typename T, typename U>
    size_t operator()(const pair<T, U> &x) const {
	size_t seed = hash<T>()(x.first);
	return hash<U>()(x.second) + 0x9e3779b9 + (seed<<6) + (seed>>2);
    }
};
const ll inf = 1e18 + 9;
const ll mod = 998244353;
const double eps = 1e-8;
const double pi = acos(-1);

int n, q;
int c[500001];
int l[500001], r[500001];

int m[500001];
int k = 0;
void add(int v) {
    if (m[c[v]] == 0) k++;
    m[c[v]]++;
}

void del(int v) {
    m[c[v]]--;
    if (m[c[v]] == 0) {
        k--;
    }
}

void solve() {
    int sq = (int)sqrt(n);
    vector<T> qi(q);
    for (int i = 0; i < q; i++) {
        qi[i] = T(l[i]/sq, r[i], i);
    }
    sort(ALL(qi));

    int nl = 0, nr = 0;
    int res[q];
    for (T& t : qi) {
        int i = get<2>(t);
        while (nr < r[i]) {
            add(nr);
            nr++;
        }
        while (nl > l[i]) {
            nl--;
            add(nl);
        }
        while (nl < l[i]) {
            del(nl);
            nl++;
        }
        while (nr > r[i]) {
            nr--;
            del(nr);
        }
        res[i] = k;
    }

    for (int i = 0; i < q; i++) {
        printf("%d\n", res[i]);
    }
}

void input() {
    scanf("%d%d", &n, &q);
    for (int i = 0; i < n; i++) scanf("%d", c+i);
    for (int i = 0; i < q; i++) {
        scanf("%d%d", l+i, r+i);
        l[i]--;
    }
}

int main() {
    input();
    solve();
}