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
from operator import itemgetter
buf = sys.stdin.buffer
input = buf.readline
class BIT:
    def __init__(self, n):
        self.n = n
        self.tree = [0] * (n + 1)
    def add(self, i, x):
        n, tree = self.n, self.tree
        while i <= n:
            tree[i] += x
            i += i & -i
    def sum(self, i):
        tree = self.tree
        s = 0
        while i:
            s += tree[i]
            i -= i & -i
        return s
def main():
    n, q = map(int, input().split())
    res = [0] * q
    *c, = map(int, input().split())
    q = [(i, l, r) for i, (l, r) in enumerate(zip(*[map(int, buf.read().split())] * 2))]
    q.sort(key=itemgetter(2))
    bit = BIT(n + 2)
    add, sum = bit.add, bit.sum
    b = [1] * (n + 1)
    j = 1
    for i, l, r in q:
        while j <= r:
            d = c[j - 1]
            j += 1
            k = b[d]
            b[d] = j
            add(k + 1, 1)
            add(j + 1, -1)
        res[i] = sum(l + 1)
    print(' '.join(map(str, res)))
main()