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
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())
c = list(map(int, input().split()))
last = [-1]*n
foradd = [[] for i in range(n)]
for i in range(n):
    if last[c[i]-1] != -1: foradd[last[c[i]-1]].append(i)
    last[c[i]-1] = i
forsum = [[] for i in range(n)]
for i in range(q):
    l, r = map(int, input().split()); l -= 1; r -= 1
    forsum[l].append([r, i])
bit = Bit(n); ans = [-1]*q
for i in range(n-1, -1, -1):
    for j in foradd[i]: bit.add(j, 1)
    for j in forsum[i]: ans[j[1]] = j[0]-i+1-bit.sum(j[0])
print(*ans, sep="\n")