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

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

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

    def sumup(self, i):
        # Sum a[1]~a[i]
        res = 0
        while i > 0:
            res += self.bit[i]
            i -= i & -i
        return res

    def query(self, i, j):
        # Sum a[i]~a[j]
        return self.sumup(j) - self.sumup(i - 1)


N, Q = map(int, input().split())
C = list(map(int, input().split()))

X = []
for i in range(Q):
    l, r = map(int, input().split())
    X.append((i, l, r))
X.sort(key=lambda x: x[2])

ans = [0] * Q
bit = BIT(N + 1)
last = [0] * (N + 1)
k = 0
for i, l, r in X:
    for j in range(k + 1, r + 1):
        if last[C[j - 1]] > 0:
            bit.add(last[C[j - 1]], -1)

        bit.add(j, 1)
        last[C[j - 1]] = j

    k = r
    ans[i] = bit.query(l, r)

print(*ans, sep="\n")
