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

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline


class BIT:
    def __init__(self, n):
        self.n = n
        self.data = [0] * (n + 1)

    def add(self, i, x=1):
        i += 1
        while i <= self.n:
            self.data[i] += x
            i += i & -i

    def get_sum(self, i):
        i += 1
        x = 0
        while i > 0:
            x += self.data[i]
            i -= i & -i
        return x

    # Return sum for [l, r)
    def get_sum_range(self, l, r):
        return self.get_sum(r - 1) - self.get_sum(l - 1)


def main():
    N, Q = map(int, readline().split())
    C = list(map(int, readline().split()))
    LR = map(int, read().split())

    prev = [-1] * (N + 1)
    ps = [[] for _ in range(N)]
    for i, c in enumerate(C):
        if prev[c] != -1:
            ps[prev[c]].append(i)
        prev[c] = i

    qs = [[] for _ in range(N)]
    for i, (l, r) in enumerate(zip(*[iter(LR)] * 2)):
        qs[l - 1].append((r - 1, i))

    ans = [0] * Q
    bit = BIT(N)

    for l in range(N - 1, -1, -1):
        for r in ps[l]:
            bit.add(r)
        for r, i in qs[l]:
            ans[i] = (r - l + 1) - bit.get_sum(r)

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

    return


if __name__ == '__main__':
    main()
