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 fenwick_tree:
    """
    区間の一点更新と，区間和の取得がO(log n)で可能なデータ構造
    1-indexedで実装
    """

    def __init__(self, N):
        self.size = N
        self.tree = [0] * (N+1)

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

    def sum_acc(self, i, j):
        """ [i,j] の和を返す """
        return self.sum_until(j) - self.sum_until(i-1)

    def add(self, i, x):
        if i <= 0:
            return
        while i <= self.size:
            self.tree[i] += x
            i += i & (-i)


def main():
    N, Q = (int(i) for i in input().split())
    C = [int(i) for i in input().split()]
    Query = [[] for _ in range(N+1)]
    for j in range(Q):
        le, ri = (int(i) for i in input().split())
        Query[ri].append((le, j))

    Query.append((-1, -1, -1))

    lastappend = [-1] * (N + 1)
    bit = fenwick_tree(N)
    ans = [0]*Q

    for i, a in enumerate(C, start=1):
        if lastappend[a] != -1:
            bit.add(lastappend[a], -1)
        lastappend[a] = i
        bit.add(i, 1)
        for (le, j) in Query[i]:
            ri = i
            ans[j] = bit.sum_acc(le, ri)

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


if __name__ == '__main__':
    main()
