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
from bisect import *
import sys
input = sys.stdin.readline
n, q = map(int, input().split())
c = list(map(int, input().split()))
class BIT():
    def __init__(self, n):
        self.size = n
        self.bit = [0] * (n+1)

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

    def add(self, i, x):
        while i <= self.size:
            self.bit[i] += x
            i += i & -i
query = [list(map(int, input().split()))+[i] for i in range(q)]
query.sort(key = lambda x: x[1])
appear = [-1]*(n+2)
bit = BIT(n)

right = 0
ans = [None]*(q)
for l, r, z in query:
    for i in range(right, r):
        num = appear[c[i]]
        if num != -1:
            bit.add(num+1, -1)
        appear[c[i]] = i
        bit.add(i+1, 1)
    right = r
    ans[z] = bit.sum(r) - bit.sum(l-1)

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