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 sys import stdin
input = stdin.readline
class Bit:
    def __init__(self, n):
        self.size = n
        self.tree = [0] * (n + 1)
 
    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())
tmp = list(map(int, input().split()))
color = [-1]
for v in tmp:
    color.append(v)
qry = [None]*q
ans = [0]*q
p = 0
for _ in ans:
    l, r = map(int, input().split())
    qry[p] = [l-1, r, p]
    p += 1
qry.sort(key=lambda x: x[1])
dic = {}
last_update_idx = 1
bit = Bit(n)
for l, r, i in qry:
    if last_update_idx < r+1:
        st = set()
        for j in range(last_update_idx, r+1):
            if not color[j] in st and color[j] in dic:
                bit.add(dic[color[j]], -1)
            dic[color[j]] = j
            st.add(color[j])
        for v in st:
            bit.add(dic[v], 1)
        last_update_idx = r+1
    ans[i] = bit.sum(r) - bit.sum(l)
for v in ans:
    print(v)