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


def bit_add(arr, n, i, x):
    while i <= n:
        arr[i] += x
        i += i & -i


def bit_sum(arr, i):
    result = 0
    while i > 0:
        result += arr[i]
        i ^= i & -i
    return result


n, q = map(int, sys.stdin.buffer.readline().split())
ccc = list(map(int, sys.stdin.buffer.readline().split()))

mp = map(int, sys.stdin.buffer.read().split())
itr = zip(mp, mp)
rli = [0] * q
for i in range(q):
    l, r = next(itr)
    rli[i] = (r << 40) + (l << 20) + i
rli.sort()
mask = (1 << 20) - 1

bn = n + 2
bit = [0] * (bn + 1)
ans = [0] * q
appeared = [-1] * (n + 1)
p = 0

for j in range(q):
    k = rli[j]
    r = k >> 40
    l = (k >> 20) & mask
    i = k & mask
    while p < r:
        c = ccc[p]
        before_appeared = appeared[c]
        if before_appeared != -1:
            bit_add(bit, bn, before_appeared + 2, -1)
        bit_add(bit, bn, p + 2, 1)
        appeared[c] = p
        p += 1

    ans[i] = bit_sum(bit, r + 1) - bit_sum(bit, l)

print('\n'.join(map(str, ans)))
