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 Python (3.8.2) (judge verdict: Time Limit Exceeded)

# Python3 code to find number of  
# distinct numbers in a subarray 
MAX = 500001
  
# structure to store queries 
class Query: 
    def __init__(self, l, r, idx): 
        self.l = l 
        self.r = r 
        self.idx = idx 
  
# updating the bit array 
def update(idx, val, bit, n): 
    while idx <= n: 
        bit[idx] += val 
        idx += idx & -idx 
  
# querying the bit array 
def query(idx, bit, n): 
    summ = 0
    while idx: 
        summ += bit[idx] 
        idx -= idx & -idx 
    return summ 
  
def answeringQueries(arr, n, queries, q): 
  
    # initialising bit array 
    bit = [0] * (n + 1) 
  
    # holds the rightmost index of  
    # any number as numbers of a[i] 
    # are less than or equal to 10^6 
    last_visit = [-1] * MAX
  
    # answer for each query 
    ans = [0] * q 
  
    query_counter = 0
    for i in range(n): 
  
        # If last visit is not -1 update -1 at the 
        # idx equal to last_visit[arr[i]] 
        if last_visit[arr[i]] != -1: 
            update(last_visit[arr[i]] + 1, -1, bit, n) 
  
        # Setting last_visit[arr[i]] as i and  
        # updating the bit array accordingly 
        last_visit[arr[i]] = i 
        update(i + 1, 1, bit, n) 
  
        # If i is equal to r of any query store answer 
        # for that query in ans[] 
        while query_counter < q and queries[query_counter].r == i: 
            ans[queries[query_counter].idx] = query(queries[query_counter].r + 1, bit, n) - query(queries[query_counter].l, bit, n) 
            query_counter += 1
  
    # print answer for each query 
    for i in range(q): 
        print(ans[i]) 
  
# Driver Code 
n,q=map(int,input().split())
a=list(map(int,input().split()))
queries=[]
for i in range(q):
  c,d=map(int,input().split())

  queries.append(Query(c-1,d-1,i))
queries.sort(key = lambda x: x.r) 
answeringQueries(a, n, queries, q)