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, heapq
input = sys.stdin.readline
n, q = map(int, input().split())
c = [int(x) - 1 for x in input().split()]

que = []
for i in range(q):
    l, r = map(int, input().split())
    que.append((l, r, i))



class SegTree(object):
    def __init__(self, n):
        self.BIT = [0] * (n + 1)
        self.num = n

    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.num:
            self.BIT[i] += x
            i += i & (-i)
    

que.sort(key=lambda x: -x[1])
ans = [None] * q
last = [-1] * n
BIT = SegTree(n)

for i in range(1, n+1):
    if last[c[i-1]] != -1:
        BIT.add(last[c[i-1]], -1)
    last[c[i-1]] = i
    BIT.add(i, 1)

    while que and que[-1][1] == i:
        l, r, x = que.pop()
        ans[x] = BIT.sum(r) - BIT.sum(l-1)

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