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.

Examples

Input

4 3
1 2 1 3
1 3
2 4
3 3


Output

2
3
1


Input

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


Output

1
2
2
1
2
2
6
3
3
3

class BIT:
    '''
    All methods use 0-based index while the implementation use 1-based index
    '''
    def __init__(self, N, val=0):
        self.data = [val] * (N + 1)
        self.N = N
    
    def add_at(self, idx, val):
        i = idx + 1
        while i <= self.N:
            self.data[i] += val
            i += i & (-i)
    
    def sum_to(self, idx):
        res = 0
        i = idx + 1
        while i > 0:
            res += self.data[i]
            i -= i & (-i)
        return res
    
    def sum_in(self, l, r): # [l, r]
        return self.sum_to(r) - self.sum_to(l - 1)
    
    
class Query:
    def __init__(self, i, l, r):
        self.i = i
        self.l = l
        self.r = r
    
    
N, Q = map(int, input().split())
C = list(map(int, input().split()))
    
queries = [[] for _ in range(N)]
for i in range(Q):
    l, r = map(int, input().split())
    l, r = l - 1, r - 1
    queries[r].append(Query(i, l, r))
    
bit = BIT(N)
last = [-1] * (max(C) + 1)
    
query_idx = 0
ans = [-1] * Q
for i, c in enumerate(C):
    if last[c] != -1:
        bit.add_at(last[c], -1)    
    bit.add_at(i, +1)
    last[c] = i
    
    for q in queries[i]:
        ans[q.i] = bit.sum_in(q.l, q.r)
    
print('\n'.join(map(str, ans)))