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
import sys

input = sys.stdin.readline

def query(i):
    '''
    a[0] + a[1] + … + a[i-1] を求める
    O(logN)
    '''
    res = 0
    while i > 0:
        res += BIT[i]
        i -= i & -i
    
    return res


def update(i, x):
    '''
    a[i-1]にxを加算
    O(logN)
    '''
    while i <= n:
        BIT[i] += x
        i += i & -i


n, q = map(int, input().split())
c = list(map(int, input().split()))
query_ = [list(map(int, input().split())) + [i] for i in range(q)]

query_.sort(key=lambda x: x[1])

res = [0] * q
pos = [-1] * n
BIT = [0] * (n+1)
p = 0
for l, r, idx in query_:
    for i in range(p, r):
        if pos[c[i]-1] != -1:
            update(pos[c[i]-1]+1, -1)
        update(i+1, 1)
        pos[c[i]-1] = i
    
    res[idx] = query(r) - query(l-1)
    p = r

print(*res, sep='\n')