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
from operator import itemgetter
import sys


class BIT:
    def __init__(self, size: int):
        self.size = size + 1
        self.array = [0] * self.size

    def add(self, i, x):
        if not i > 0:
            return
        while i < self.size:
            self.array[i] += x
            i += i & -i

    def sum(self, i):
        s = 0
        while i > 0:
            s += self.array[i]
            i -= i & -i
        return s


def read():
    return sys.stdin.readline().rstrip()


def main():
    n, q = map(int, read().split())
    c = [int(i) for i in read().split()]
    queries = [() for _ in range(q)]
    for i in range(q):
        l, r = map(int, read().split())
        queries[i] = (i, l, r)
    queries.sort(key=itemgetter(2))
    last = [0] * n
    bit = BIT(n)
    ans = [-1] * q
    cr = 0
    for i, l, r in queries:
        for ri in range(cr, r):
            ci = c[ri] - 1
            bit.add(last[ci], -1)
            last[ci] = ri + 1
            bit.add(last[ci], 1)
            cr = ri
        ans[i] = bit.sum(r) - bit.sum(l - 1)
    print(*ans, sep="\n")


if __name__ == '__main__':
    main()
