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)

#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
#define endl "\n"
#define sz(x) ((ll)(x).size())
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define prt(x) cout << (x)
#define prtl(x) cout << (x) << endl
#define rep(i,a,b) for(ll i=a; i<b; i++)
#define rrep(i,a,b) for(ll i=a; i>b; i--)
#define mp(a, b) make_pair(a, b)
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define debug(v) cout << #v<< ": " << v <<endl;
#define fst first
#define snd second
using namespace std;
using vi = vector<int>;
using vll = vector<long long int>;
using vb = vector<bool>;
using vvb = vector<vector<bool>>;
using vc = vector<char>;
using vvi = vector<vector<int>>;
using vvll = vector<vector<long long int>>;
using vvc = vector<vector<char>>;
using ll = long long int;
using P = pair<long long int, long long int>;
using Map = map<long long int, long long int>;
ll INF = 1LL<<60;
ll M = 1000000007;

struct Mo
{
  vector< int > left, right, order;
  vector< bool > v;
  int width;
  int nl, nr, ptr;

  Mo(int n) : width((int) sqrt(n)), nl(0), nr(0), ptr(0), v(n) {}

  void insert(int l, int r) /* [l, r) */
  {
    left.push_back(l);
    right.push_back(r);
  }

  /* ソート */
  void build()
  {
    order.resize(left.size());
    iota(begin(order), end(order), 0);
    sort(begin(order), end(order), [&](int a, int b)
    {
      if(left[a] / width != left[b] / width) return left[a] < left[b];
      return right[a] < right[b];
    });
  }

  /* クエリを 1 つぶんすすめて, クエリのidを返す */
  int process()
  {
    if(ptr == order.size()) return (-1);
    const auto id = order[ptr];
    while(nl > left[id]) distribute(--nl);
    while(nr < right[id]) distribute(nr++);
    while(nl < left[id]) distribute(nl++);
    while(nr > right[id]) distribute(--nr);
    return (order[ptr++]);
  }

  inline void distribute(int idx)
  {
    v[idx].flip();
    if(v[idx]) add(idx);
    else del(idx);
  }

  void add(int idx);

  void del(int idx);
};

int N, A[500001], Q;
int ans[500001];
int cnt[500001], sum;

void Mo::add(int idx)
{
  if(cnt[A[idx]]++ == 0) ++sum;
}

void Mo::del(int idx)
{
  if(--cnt[A[idx]] == 0) --sum;
}

int main()
{
  cin >> N;
  cin >> Q;
  for(int i = 0; i < N; i++) {
    cin >> A[i];
  }
  Mo mo(N);
  for(int i = 0; i < Q; i++) {
    int a, b;
    cin >> a >> b;
    mo.insert(--a, b);
  }
  mo.build();
  for(int i = 0; i < Q; i++) {
      int g = mo.process();
       ans[g] = sum;
  }
  for(int i = 0; i < Q; i++) {
    prtl(ans[i]);
  }
}