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
def solve():
    import sys
    input = sys.stdin.readline

    class BIT:
        def __init__(self, n):
            self.n = n
            self.data = [0] * n

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

        def update(self, i, x):
            while i < self.n:
                self.data[i] += x
                i |= i + 1

    n, q = map(int, input().split())
    c = list(map(int, input().split()))
    odr_r = [[] for _ in range(n + 1)]
    for i in range(q):
        l, r = map(int, input().split())
        odr_r[r].append((l - 1, i))

    b = BIT(n)
    diff = 0
    ans = [-1] * q
    last = [-1] * (n + 1)

    for i, v in enumerate(c):
        if last[v] != -1:
            b.update(last[v], -1)
        else:
            diff += 1

        b.update(i, 1)
        last[v] = i

        for l, j in odr_r[i + 1]:
            ans[j] = diff - b.sum(l)
    print("\n".join(map(str, ans)))
    return


solve()
