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 BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>
#define FASTIO ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define rep(i, l, r) for (ll i = (l); i < (r); ++i)
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define pii pair<ll, ll>
#define pll pair<ll, ll>
#define pli pair<ll, ll> 
#define pdd pair<ld, ld> 
#define pb push_back
#define reset(x,v) memset(x,v,sizeof(x))
#define mod 1000000007
using namespace std;
set<int>* segment; 
  
void build(int i, int s, int e, int arr[]) 
{ 

    if (s == e) { 
        segment[i].insert(arr[s]); 
        return; 
    } 
  

    build(2 * i, s, (s + e) / 2, arr); 
    build(1 + 2 * i, 1 + (s + e) / 2, e, arr); 
  
    segment[i].insert(segment[2 * i].begin(), 
                      segment[2 * i].end()); 
  
    segment[i].insert(segment[2 * i + 1].begin(), 
                      segment[2 * i + 1].end()); 
} 
  
set<int> query(int node, int l, int r, int a, int b) 
{ 
    set<int> left, right, result; 
  
    if (b < l || a > r) 
        return result; 

    if (a <= l && r <= b) 
        return segment[node]; 
  
    left = query(2 * node, l, (l + r) / 2, a, b); 
    result.insert(left.begin(), left.end()); 
  
    right = query(1 + 2 * node, 1 + (l + r) / 2, r, a, b); 
    result.insert(right.begin(), right.end()); 
  

    return result; 
} 
  
void init(int n) 
{ 
    int h = (int)ceil(log2(n)); 
    h = (2 * (pow(2, h))) - 1; 

    segment = new set<int>[h]; 
} 

void getDistinct(int l, int r, int n) 
{ 
    set<int> ans = query(1, 0, n - 1, l, r); 
  
    cout << ans.size() << endl; 
}
int main(){
	FASTIO
	int n, k;
	cin >> n >> k;
	int a[n];
	rep(i, 0, n)
		cin >> a[i];
	init(n+2);
	build(1, 0, n - 1, a); 
	while(k--){
		int l, r;
		cin >> l >> r;
		getDistinct(l-1, r-1, n);
	}
    return 0;
}