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
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__

class Binary_Indexed_Tree_RAQ():
    def __init__(self, n):
        self.bit = [0]*(n+1)
        self.n = n+1

    def __add(self, ind, num):
        while(ind <= self.n):
            self.bit[ind - 1] += num
            ind += ind & -ind

    def range_add(self, s, t, x):
        self.__add(s,x)
        self.__add(t+1,-x)

    def query(self, i):
        ret = 0
        while(i > 0):
            ret += self.bit[i-1]
            i -= i & -i
        return ret

n,q = map(int, input().split())
a = list(map(int, input().split()))
m = 10**6 + 1
prev = [-1]*m

query = []
for i in range(q):
    l,r = map(int, input().split())
    l,r = l-1,r-1
    query.append((l,r,i))
ans = [0]*q
query.sort(key=lambda x: x[1])

BIT = Binary_Indexed_Tree_RAQ(n+1)
cur = 0
for l,r,i in query:
    while cur <= r:
        ai = a[cur]
        BIT.range_add(prev[ai]+2, cur+1, 1)
        prev[ai] = cur
        cur += 1
    ans[i] = BIT.query(l+1)
print(*ans, sep="\n")