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: Runtime Error)

// C++ implementation of above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Each segment of the segment tree would be a set 
// to maintain distinct elements 
set<int>* segment; 
  
// Build the segment tree 
// i denotes current node, s denotes start and 
// e denotes the end of range for current node 
void build(int i, int s, int e, int arr[]) 
{ 
  
    // If start is equal to end then 
    // insert the array element 
    if (s == e) { 
        segment[i].insert(arr[s]); 
        return; 
    } 
  
    // Else divide the range into two halves 
    // (start to mid) and (mid+1 to end) 
    // first half will be the left node 
    // and the second half will be the right node 
    build(2 * i, s, (s + e) / 2, arr); 
    build(1 + 2 * i, 1 + (s + e) / 2, e, arr); 
  
    // Insert the sets of right and left 
    // node of the segment tree 
    segment[i].insert(segment[2 * i].begin(), 
                      segment[2 * i].end()); 
  
    segment[i].insert(segment[2 * i + 1].begin(), 
                      segment[2 * i + 1].end()); 
} 
  
// Query in an range a to b 
set<int> query(int node, int l, int r, int a, int b) 
{ 
    set<int> left, right, result; 
  
    // If the range is out of the bounds 
    // of this segment 
    if (b < l || a > r) 
        return result; 
  
    // If the range lies in this segment 
    if (a <= l && r <= b) 
        return segment[node]; 
  
    // Else query for the right and left 
    // leaf node of this subtree 
    // and insert them into the set 
    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 the result 
    return result; 
} 
  
// Initialize the segment tree 
void init(int n) 
{ 
    // Get the height of the segment tree 
    int h = (int)ceil(log2(n)); 
    h = (2 * (pow(2, h))) - 1; 
  
    // Initialize the segment tree 
    segment = new set<int>[h]; 
} 
  
// Function to get the result for the 
// subarray from arr[l] to arr[r] 
void getDistinct(int l, int r, int n) 
{ 
    // Query for the range set 
    set<int> ans = query(1, 0, n - 1, l, r); 
  
    cout << ans.size() << endl; 
} 
  
// Driver code 
int main() 
{ 
	int n,q;
	cin>>n>>q;
	int arr[n],i;
	for(i=0;i<n;i++)
		cin>>arr[i];
  
    // int arr[] = { 1, 1, 2, 1, 3 }; 
    // int n = sizeof(arr) / sizeof(arr[0]); 
  
    init(n); 
  
    // Bulid the segment tree 
    build(1, 0, n - 1, arr); 
  
    // Query in range 0 to 4 
    for(i=0;i<q;i++)
    {
    	int l,r;
    	cin>>l>>r;
    	l--;
    	r--;
    	getDistinct(l,r,n);
    }
    // getDistinct(0, 4, n); 
  
    return 0; 
}