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

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

    def sum(self, i):
        i += 1
        s = 0
        while i > 0:
            s += self.bit[i]
            i -= i & -i
        return s

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

    def lower_bound(self, x):
        if x <= 0:
            return 0
        k = 1
        while k * 2 <= self.size:
            k *= 2
        j = 0
        while k > 0:
            if j + k <= self.size and self.bit[j + k] < x:
                x -= self.bit[j + k]
                j += k
            k //= 2
        return j

N, Q = map(int, input().split())
c = [0] + list(map(int, input().split()))
queries = [list(map(int, input().split())) + [i] for i in range(Q)]
queries.sort(key=lambda q: q[1])
last = [-1] * (N+1)
ans = [0] * Q
bit = BIT(N + 1)
j = 0
for l, r, i in queries:
    while j < r:
        j += 1
        if last[c[j]] != -1:
            bit.add(last[c[j]], -1)
        bit.add(j, 1)
        last[c[j]] = j
    ans[i] = bit.sum(r) - bit.sum(l - 1)
print(*ans, sep='\n')
