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

    def query(self, i):
        ret = 0
        i += 1
        while i:
            ret += self.arr[i]
            lsb = i & (-i)
            i -= lsb
        return ret

    def add(self, i, v):
        i += 1
        while i < self.N+1:
            lsb = i & (-i)
            self.arr[i] += v
            i += lsb


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

    queries = [None] * Q
    for i in range(Q):
        l, r = map(int, input().split())
        queries[i] = [l, r, i]
    queries.sort(key=lambda x: x[1])

    bit = BinaryIndexedTree(N)
    numdupls = 0
    prevs = [-1] * N
    cur = 0
    ans = [-999] * Q
    for i in range(Q):
        l, r, qindex = queries[i]
        while cur < r:
            c = C[cur] - 1
            if prevs[c] != -1:
                bit.add(prevs[c], 1)
                numdupls += 1
            prevs[c] = cur
            cur += 1
        ans[qindex] = r-l+1 - (numdupls - bit.query(l-2))
#        print(ans, "l=", l, "r=", r, numdupls, bit.query(l-2))
    for i in range(Q):
        print(ans[i])


if __name__ == "__main__":
    main()
