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
import sys
input = sys.stdin.readline


class BIT:
    def __init__(self, a):
        self.n = len(a)
        self.a = [0] * (self.n + 1)

        for i, j in enumerate(a):
            self.update(i, j)

    def update(self, i, x):
        i += 1
        while i <= self.n:
            self.a[i] = self.a[i] + x
            i += i & (-i)

    def get(self, i):
        s = 0
        i += 1
        while i != 0:
            s = s + self.a[i]
            i -= i & (-i)
        return s


def inmin(x):
    return int(x)-1


if __name__ == "__main__":
    n, q = map(int, input().split())
    if n>5*10**5 or q>5*10**5:
        a=1/0
    c = list(map(inmin, input().split()))
    que = [[i, list(map(inmin, input().split()))] for i in range(q)]
    que.sort(key=lambda x: x[1][1])
    ans = [0] * q

    last = [-1] * n
    bit = BIT([0]*(n+10))
    j = 0
    for i in range(n):
        if last[c[i]] >= 0:
            bit.update(last[c[i]]+5, -1)
        bit.update(i+5, 1)
        last[c[i]] = i
        while j < q and que[j][1][1] == i:
            ans[que[j][0]] = bit.get(i+5) - bit.get(que[j][1][0]+4)
            j += 1
    print(*ans)
