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
SHIFT = 1 << 20


def main():
    N, Q = (int(i) for i in input().split())
    C = [int(i) for i in input().split()]
    Query = []
    for j in range(Q):
        le, ri = (int(i) for i in input().split())
        Query.append(ri << 40 | le << 20 | j)
    Query.sort()

    # Query復元
    for i in range(Q):
        j = Query[i] % (SHIFT)
        Query[i] >>= 20
        le = Query[i] % (SHIFT)
        Query[i] >>= 20
        ri = Query[i]
        Query[i] = (le, ri, j)
    Query.append((-1, -1, -1))

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

    def add(tree, i, x):
        while i <= N:
            tree[i] += x
            i += i & -i

    lastappend = [-1] * (N + 1)
    bit = [0] * (N + 1)
    idx = 0  # Queryのindex
    ans = [0]*Q

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

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


if __name__ == '__main__':
    main()
