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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rep1(i, n) FOR(i, 1, n+1)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; --i)
#define whole(x) (x).begin(),(x).end()
#define rwhole(x) (x).rbegin(), (x).rend()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define P pair<int, int>
#define debug(var) cerr << "[" << #var << "] " << var << endl
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define eb emplace_back
const ll mod = 1000000007;
const int dx[] = {-1,0,1,0};
const int dy[] = {0,-1,0,1};
 
struct Query {
    int l, r, i;
    Query(int l, int r, int i): l(l), r(r), i(i){}
};
 
template<typename T>
struct BIT {
  int n;
  vector<T> d;
  BIT(int n=0):n(n),d(n+1) {}
  void add(int i, T x=1) {
    for (i++; i <= n; i += i&-i) {
      d[i] += x;
    }
  }
  T sum(int i) {
    T x = 0;
    for (i++; i; i -= i&-i) {
      x += d[i];
    }
    return x;
  }
  T sum(int l, int r) {
    return sum(r-1) - sum(l-1);
  }
};
 
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
    int n, q;
    cin >> n >> q;
    vector<int> c(n);
    rep(i, n) cin >> c[i];
    vector<Query> query;
    rep(qi, q) {
        int l, r;
        cin >> l >> r;
        l--; r--;
        query.eb(l, r, qi);
    }
    auto comp = [](Query a, Query b) {
        return a.l<b.l;
    };
    sort(whole(query), comp);
    reverse(whole(query));
    vector<int> ans(q);
    vector<int> pi(n+1, -1);
    vector<P> p;
    rep(i, n) {
        int prev = pi[c[i]];
        if (prev!=-1) {
            p.emplace_back(prev, i);
        }
        pi[c[i]] = i;
    }
    BIT<int> bit(n+5);
    sort(whole(p));
    reverse(whole(p));
    int index = 0;
 
    rep(i, q) {
        int l = query[i].l;
        int r = query[i].r;
        int qi = query[i].i;
        while (index<p.size()) {
            if (p[index].first<l) break;
            bit.add(p[index].second, 1);
            index++;
        }
        int a = bit.sum(r);
        ans[qi] = r - l + 1- a;
    }
 
    rep(i, q) cout << ans[i] << "\n";
    return 0;
}