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.readline
N,Q = map(int,input().split())
C = list(map(int,input().split()))
LR = [tuple(map(int,input().split())) for i in range(Q)]

qs = [[] for _ in range(N)]
for i,(l,r) in enumerate(LR):
    qs[l-1].append((i,r-1))

from collections import defaultdict
hist = defaultdict(lambda: -1)
nx = [-1] * N
for i in range(N-1,-1,-1):
    nx[i] = hist[C[i]]
    hist[C[i]] = i

class BinaryIndexedTree:
    def __init__(self,size):
        self.N = size
        self.bit = [0]*(size+1)
    def add(self,x,w): # 0-indexed
        x += 1
        while x <= self.N:
            self.bit[x] += w
            x += (x & -x)
    def _sum(self,x): # 1-indexed
        ret = 0
        while x > 0:
            ret += self.bit[x]
            x -= (x & -x)
        return ret
    def sum(self,l,r): # [l,r)
        return self._sum(r) - self._sum(l)
    def __str__(self): # for debug
        arr = [self.sum(i,i+1) for i in range(self.N)]
        return str(arr)
bit = BinaryIndexedTree(N)

ans = [None] * Q
for l in range(N-1,-1,-1):
    if nx[l] != -1:
        bit.add(nx[l], -1)
    for i,r in qs[l]:
        ans[i] = r-l+1 + bit.sum(l,r+1)

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