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.buffer.readline
n, q = map(int, input().split())
c = list(map(lambda x:int(x)-1, input().split()))
query = []
for i in range(q):
    l, r = map(int, input().split())
    query.append(i+(l-1)*10**6+(r-1)*10**12)

class BinaryIndexedTree():
    def __init__(self, n):
        self.n = n
        self.bit = [0]*n
    
    def add(self, i, w):
        x = i+1
        while x<=self.n:
            self.bit[x-1] += w
            x += x & -x
    
    def sum(self, i):
        ret = 0
        x = i+1
        while x>0:
            ret += self.bit[x-1]
            x -= x & -x
        return ret

bit = BinaryIndexedTree(n+1)
query.sort()
last = [None]*n
ans = [None]*q
nr = -1
for i in query:
    r = i//10**12
    l = i%10**12//10**6
    idx = i%10**6
    for cr in range(nr+1, r+1):
        if last[c[cr]] is not None:
            bit.add(last[c[cr]]+1, -1)
        last[c[cr]] = cr
        bit.add(cr+1, 1)
    nr = r
    ans[idx] = bit.sum(r+1) - bit.sum(l)
print(*ans)