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)

/*
このコード、と～おれ!
Be accepted!
∧＿∧　
（｡･ω･｡)つ━☆・*。
⊂　　 ノ 　　　・゜+.
　しーＪ　　　°。+ *´¨)
 　　　　　　　　　.· ´¸.·*´¨) ¸.·*¨)
		  　　　　　　　　　　(¸.·´ (¸.·'* ☆
*/

#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <vector>
#include <numeric>
#include <iostream>
#include <random>
#include <map>
#include <unordered_map>
#include <queue>
#include <regex>
#include <functional>
#include <complex>
#include <list>
#include <cassert>
#include <iomanip>
#include <set>
#include <stack>
#include <bitset>

////多倍長整数, cpp_intで宣言
//#include <boost/multiprecision/cpp_int.hpp>
//using namespace boost::multiprecision;

//#pragma GCC target ("avx2")
#pragma GCC optimization ("Ofast")
#pragma GCC optimization ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define repeat(i, n, m) for(int i = n; i < (m); ++i)
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define printynl(a) printf(a ? "yes\n" : "no\n")
#define printyn(a) printf(a ? "Yes\n" : "No\n")
#define printYN(a) printf(a ? "YES\n" : "NO\n")
#define printim(a) printf(a ? "possible\n" : "imposible\n")
#define printdb(a) printf("%.50lf\n", a) //少数出力
#define printLdb(a) printf("%.50Lf\n", a) //少数出力
#define printdbd(a) printf("%.16lf\n", a) //少数出力(桁少なめ)
#define prints(s) printf("%s\n", s.c_str()) //string出力
#define all(x) (x).begin(), (x).end()
#define deg_to_rad(deg) (((deg)/360.0L)*2.0L*PI)
#define rad_to_deg(rad) (((rad)/2.0L/PI)*360.0L)
#define Please return
#define AC 0
#define manhattan_dist(a, b, c, d) (abs(a - c) + abs(b - d)) /*(a, b) から (c, d) のマンハッタン距離 */
#define inf numeric_limits<double>::infinity();
#define linf numeric_limits<long double>::infinity()

using ll = long long;
using ull = unsigned long long;

constexpr int INF = 1073741823;
constexpr int MINF = -1073741823;
constexpr ll LINF = ll(4661686018427387903);
constexpr ll MOD = 1e9 + 7;
constexpr ll mod = 998244353;
constexpr long double eps = 1e-6;
const long double PI = acosl(-1.0L);

using namespace std;

void scans(string& str) {
	char c;
	str = "";
	scanf("%c", &c);
	if (c == '\n')scanf("%c", &c);
	while (c != '\n' && c != -1 && c != ' ') {
		str += c;
		scanf("%c", &c);
	}
}

void scanc(char& str) {
	char c;
	scanf("%c", &c);
	if (c == -1)return;
	while (c == '\n') {
		scanf("%c", &c);
	}
	str = c;
}


double acot(double x) {
	return PI / 2 - atan(x);
}

ll LSB(ll n) { return (n & (-n)); }

template<typename T>
inline T chmin(T& a, const T& b) {
	if (a > b)a = b;
	return a;
}

template<typename T>
inline T chmax(T& a, const T& b) {
	if (a < b)a = b;
	return a;
}

/*-----------------------------------------ここからコード-----------------------------------------*/

/*
* @title Mo's Algorithm
* @docs kyopro/docs/mo.md
*/

template<typename ADD, typename DEL, typename REM, typename T = int>
struct mo {
	int sqn, q, l, r, p;
	T ret;
	vector<tuple<int, int, int>> query;
	vector<T> ans;

	mo(const int& n, const int& q) : sqn((int)sqrt(n)), q(q), l(0), r(0), p(0), ret(T(0)), query(q), ans(q) {}

	inline void insert(const int& l, const int& r) {
		query[p] = { l, r, p++ };
	}

	inline void read(const bool& oneindexed) {
		for (auto& [left, right, idx] : query) {
			scanf("%d%d", &left, &right);
			if (oneindexed)--left;
			idx = p++;
		}
	}

	void build() {
		sort(all(query), [&](const tuple<int, int, int>& a, const tuple<int, int, int>& b) {
			if (get<0>(a) / sqn != get<0>(b) / sqn)return get<0>(a) < get<0>(b);
			return get<1>(a) < get<1>(b);
			});
	}

	inline void run(const ADD& add, const DEL& del, const REM& rem) {
		for (const auto& [ql, qr, qo] : query) {
			while (l > ql)add(--l, ret);
			while (r < qr)add(r++, ret);
			while (l < ql)del(l++, ret);
			while (r > qr)del(--r, ret);
			rem(qo, ans, ret);
		}
	}

	T operator [](const int& idx) {
		return ans[idx];
	}

	void allrun(const bool& oneindexed, const ADD& add, const DEL& del, const REM& rem) {
		read(oneindexed);
		build();
		run(add, del, rem);
	}

};

inline auto rem = [](const int& idx, vector<int>& ans, const int& ret) {ans[idx] = ret; };

int main() {

	int n, q;
	scanf("%d%d", &n, &q);
	vector<int> a(n);
	rep(i, n) scanf("%d", &a[i]);
	vector<int> memo(n + 1);
	auto add = [&](const int& idx, int& ret) {if (memo[a[idx]]++ == 0)++ret; };
	auto del = [&](const int& idx, int& ret) {if (--memo[a[idx]] == 0)--ret; };
	mo<decltype(add), decltype(del), decltype(rem), int> m(n, q);
	m.allrun(true, add, del, rem);
	rep(i, q)printf("%d\n", m[i]);

	Please AC;
}