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 rep(i,n)for(int i=0;i<(n);i++)
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define pb push_back
using ll = long long;
using P = pair<int, int>;

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(){
    int n, q; 
    cin >> n >> q;
    vector<int> a(n);
    rep(i,n) cin >> a[i];
    vector<int> pi(n+1, -1);//a[i]が前に出てきたindex
    vector<vector<int>> ps(n);
    rep(i,n){
        int l = pi[a[i]];
        if(l != -1) ps[l].push_back(i);
        pi[a[i]] = i;
    }

    vector<vector<P>> querys(n);
    rep(qi,q){
        int l, r;
        cin >> l >> r;
        --l, --r;
        querys[l].emplace_back(r, qi);
    }
    vector<int> ans(q);
    BIT<int> d(n);
    for(int x = n-1; x >= 0; --x){
        //x左端, r右端
        for(int y : ps[x]) d.add(y, 1);
        for(P q : querys[x]){
            int r = q.first, i = q.second;
            ans[i] = (r - x + 1) - d.sum(r);
        }
    }
    rep(i,q) cout << ans[i] << endl;
}