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)

#ifndef ONLINE_JUDGE
#define DEBUG 1
#if DEBUG
#define _GLIBCXX_DEBUG
#endif
#endif
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstring>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <utility>
#include <vector>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using vll = vector<ll>;
using vvll = vector<vll>;
using pll = pair<ll, ll>;
using vpll = vector<pll>;
using vvpll = vector<vpll>;
using tll = tuple<ll, ll, ll>;
using vtll = vector<tll>;
using vvtll = vector<vtll>;
#define all(v) (v).begin(), (v).end()
#define for1(i, n) for (ll i = 0; i < (n); i++)
#define for2(i, m, n) for (ll i = (m); i < (n); i++)
#define for3(i, m, n, d) for (ll i = (m); i < (n); i += (d))
#define rfor2(i, m, n) for (ll i = (m); i > (n); i--)
#define rfor3(i, m, n, d) for (ll i = (m); i > (n); i += (d))
#define PI 3.1415926535897932384626433832795028841971693993751L
#define INF 1111111111111111111LL
#define print(...) print_1(__VA_ARGS__)
#define in(...) in_1(__VA_ARGS__)
#if DEBUG
#define dump(...) dump_1(#__VA_ARGS__, __VA_ARGS__)
#define dumpa(...) dumpa_1(#__VA_ARGS__, __VA_ARGS__)
#else
#define dump(...)
#define dumpa(...)
#endif
template <typename Head>
void dump_1(const char* str, Head&& h)
{
    cerr << str << ": " << h << '\n';
}
template <typename Head, typename... Tail>
void dump_1(const char* str, Head&& h, Tail&&... t)
{
    while (*str != ',') {
        cerr << *str++;
    }
    cerr << ": " << h << ' ';
    dump_1(str + 1, t...);
}
template <typename T>
void dumpa_1(const char* str, const T v[], const ll size)
{
    while (*str != ',') {
        cerr << *str++;
    }
    cerr << ": ";
    for1 (i, size) {
        if (i != 0) {
            cerr << ' ';
        }
        cerr << v[i];
    }
    cerr << '\n';
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& v)
{
    os << v.first << ' ' << v.second;
    return os;
}
template <typename T1, typename T2, typename T3>
ostream& operator<<(ostream& os, const tuple<T1, T2, T3>& v)
{
    os << get<0>(v) << ' ' << get<1>(v) << ' ' << get<2>(v);
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ' ';
        }
        os << *it;
    }
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v)
{
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ' ';
        }
        os << *it;
    }
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v)
{
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ' ';
        }
        os << *it;
    }
    return os;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const map<T1, T2>& v)
{
    os << '{';
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ", ";
        }
        os << it->first << ':' << it->second;
    }
    os << '}';
    return os;
}
ll divup(ll nume, ll deno)
{
    assert(nume >= 0);
    assert(deno > 0);
    return (nume + deno - 1) / deno;
}
void Yes(void) { cout << "Yes\n"; }
void No(void) { cout << "No\n"; }
void YES(void) { cout << "YES\n"; }
void NO(void) { cout << "NO\n"; }
template <typename T>
bool chmax(T& a, const T& b)
{
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template <typename T>
bool chmin(T& a, const T& b)
{
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <typename T>
void vin(vector<T>& v)
{
    ll len = v.size();
    for1 (i, len) {
        cin >> v[i];
    }
}
template <typename Head>
void in_1(Head& h)
{
    cin >> h;
}
template <typename Head, typename... Tail>
void in_1(Head& h, Tail&... t)
{
    cin >> h;
    in_1(t...);
}
template <typename Head>
void print_1(Head&& h)
{
    cout << h << '\n';
}
template <typename Head, typename... Tail>
void print_1(Head&& h, Tail&&... t)
{
    cout << h << ' ';
    print_1(t...);
}
//---------------------------------------------------------
const ll mod = 1000000007LL;  // 10**9 + 7
struct mint {
    ll x;  // typedef long long ll;
    mint(ll x = 0) : x((x % mod + mod) % mod) {}
    mint& operator+=(const mint a)
    {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a)
    {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a)
    {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint a) const
    {
        mint res(*this);
        return res += a;
    }
    mint operator-(const mint a) const
    {
        mint res(*this);
        return res -= a;
    }
    mint operator*(const mint a) const
    {
        mint res(*this);
        return res *= a;
    }
    mint pow(ll t) const
    {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const { return pow(mod - 2); }
    mint& operator/=(const mint a) { return (*this) *= a.inv(); }
    mint operator/(const mint a) const
    {
        mint res(*this);
        return res /= a;
    }
    friend istream& operator>>(istream& is, mint& v)
    {
        is >> v.x;
        return is;
    }
    friend ostream& operator<<(ostream& os, const mint& v)
    {
        os << v.x;
        return os;
    }
};
//---------------------------------------------------------
struct mintcomb {
    vector<mint> fact, ifact;
    mintcomb(int n) : fact(n + 1), ifact(n + 1)
    {
        assert(n < mod);
        fact[0] = 1;
        for (int i = 1; i <= n; ++i) {
            fact[i] = fact[i - 1] * i;
        }
        ifact[n] = fact[n].inv();
        for (int i = n; i >= 1; --i) {
            ifact[i - 1] = ifact[i] * i;
        }
    }
    mint permutation(int n, int k)
    {
        if (k < 0 || k > n) return 0;
        return fact[n] * ifact[n - k];
    }
    mint combination(int n, int k)
    {
        if (k < 0 || k > n) return 0;
        return fact[n] * ifact[k] * ifact[n - k];
    }
};
//---------------------------------------------------------

template <typename T>
struct RangeSetQuery {
    vector<T> value;  // ノードの値を持つ配列
    int N;            // 葉の数
    explicit RangeSetQuery(int size)
    {
        N = 1;
        while (N < size) N *= 2;  // 葉の数を計算（n以上の最小の2冪数）
        value = vector<T>(2 * N - 1);
    }

    void update(int i, T& x)
    {
        // i 番目の葉の値を x に変える
        i += N - 1;  // i 番目の葉のノード番号
        value[i] = x;
        while (i > 0) {
            // ノード i の親ノードの番号に変える
            i = (i - 1) / 2;
            // 左右の子の min を計算しなおす
            T& c1 = value[i * 2 + 1];
            T& c2 = value[i * 2 + 2];
            set_union(c1.begin(), c1.end(), c2.begin(), c2.end(),
                      inserter(value[i], value[i].end()));
        }
    }

    T query2(int a, int b, int k, int l, int r)
    {
        // [a, b) の区間に対するクエリについて
        // ノード k （区間 [l, r) 担当）が答える
        T result;
        if (r <= a || b <= l) return result;  // 区間が被らない場合は INF を返す
        if (a <= l && r <= b)
            // ノード k の担当範囲がクエリ区間 [a, b) に完全に含まれる
            return value[k];
        else {
            // 左の子に値を聞く
            T&& c1 = query2(a, b, 2 * k + 1, l, (l + r) / 2);
            // 右の子に値を聞く
            T&& c2 = query2(a, b, 2 * k + 2, (l + r) / 2, r);
            // 左右の子の値の min を取る
            set_union(c1.begin(), c1.end(), c2.begin(), c2.end(),
                      inserter(result, result.end()));
            return result;
        }
    }
    T query(int a, int b)
    {
        // [a, b) の区間に対するクエリを返す
        return query2(a, b, 0, 0, N);
    }
};

void solve()
{
    ll N, Q;
    in(N, Q);
    vll C(N);
    vin(C);
    RangeSetQuery<set<ll>> rsq(N);
    for1 (i, N) {
        set<ll> value;
        value.insert(C[i]);
        rsq.update(i, value);
    }
    vll ans(Q);
    for1 (i, Q) {
        ll l, r;
        in(l, r);
        l--;
        r--;
        auto result = rsq.query(l, r + 1);
        print(result.size());
    }
}
//---------------------------------------------------------
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(16);
    cerr << fixed << setprecision(16);
    solve();
}