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, n):
        self.size = n
        self.tree = [0 for _ in range(n + 2)]
 
    def sum(self, i):
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s
 
    def add(self, i, x):
        while i <= self.size:
            self.tree[i] += x
            i += i & -i


n, q = map(int, input().split())
c = list(map(int, input().split()))

rli = []
for i in range(q):
    l, r = map(int, input().split())
    rli.append((r, l, i))
rli.sort(key=lambda x: x[0])

last = [-1 for _ in range(n + 1)]

bit = Bit(n)
ans = [-1 for _ in range(q)]
cur = 1
for r, l, i in rli:
    for j in range(cur, r + 1):
        if last[c[j - 1]] != -1:
            bit.add(last[c[j - 1]], -1)
        last[c[j - 1]] = j
        bit.add(j, 1)
    cur = r + 1
    ans[i] = bit.sum(r) - bit.sum(l - 1)
print(*ans, sep="\n")
