One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.

If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.

To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.

Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.

We remind you that 1 byte contains 8 bits.

k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.

Input

The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.

The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.

Output

Print a single integer — the minimal possible number of changed elements.

Examples

Input


6 1
2 1 2 3 4 3


Output


2


Input


6 2
2 1 2 3 4 3


Output


0


Input


6 1
1 1 2 2 3 3


Output


2

Note

In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.

In the second example the disk is larger, so the initial file fits it and no changes are required.

In the third example we have to change both 1s or both 3s.
import sys
import os
from io import BytesIO, IOBase

#########################
# imgur.com/Pkt7iIf.png #
#########################

# returns the list of prime numbers less than or equal to n:
'''def sieve(n):
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r'''

# returns all the divisors of a number n(takes an additional parameter start):
'''def divs(n, start=1):
    divisors = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if n % i == 0:
            if n / i == i:
                divisors.append(i)
            else:
                divisors.extend([i, n // i])
    return divisors'''

# returns the number of factors of a given number if a primes list is given:
'''def divn(n, primes):
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t
    return divs_number'''

# returns the leftmost and rightmost positions of x in a given list d(if x isnot present then returns (-1,-1)):
'''def flin(d, x, default=-1):
    left = right = -1
    for i in range(len(d)):
        if d[i] == x:
            if left == -1: left = i
            right = i
    if left == -1:
        return (default, default)
    else:
        return (left, right)'''

# returns (b**p)%m 
'''def modpow(b,p,m):
    res=1
    b=b%m
    while(p):
        if p&1:
            res=(res*b)%m
        b=(b*b)%m
        p>>=1
    return res'''

# if m is a prime this can be used to determine (1//a)%m or modular multiplicative inverse of a number a
'''def mod_inv(a,m):
    return modpow(a,m-2,m)'''

# returns the ncr%m for (if m is a prime number) for very large n and r
'''def ncr(n,r,m):
    res=1
    if r==0:
        return 1
    if n-r<r:
        r=n-r
    p,k=1,1
    while(r):
        res=((res%m)*(((n%m)*mod_inv(r,m))%m))%m
        n-=1
        r-=1
    return res'''

# returns ncr%m (if m is a prime number and there should be a list fact which stores the factorial values upto n):
'''def ncrlis(n,r,m):
    return (fact[n]*(mod_inv(fact[r],m)*mod_inv(fact[n-r],m))%m)%m'''

#factorial list which stores factorial of values upto n:
'''mx_lmt=10**5+10
fact=[1 for i in range(mx_lmt)]
for i in range(1,mx_lmt):
    fact[i]=(i*fact[i-1])%mod'''

#count xor of numbers from 1 to n:
'''def xor1_n(n):
    d={0:n,1:1,2:n+1,3:0}
    return d[n&3]'''

def cel(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def prr(a, sep=' '): print(sep.join(map(str, a)))
def dd(): return defaultdict(int)
def ddl(): return defaultdict(list)
def ddd(): return defaultdict(defaultdict(int))
 

# region fastio
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
	newlines = 0
 
	def __init__(self, file):
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = "x" in file.mode or "r" not in file.mode
		self.write = self.buffer.write if self.writable else None
 
	def read(self):
			while True:
					b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
					if not b:
							break
					ptr = self.buffer.tell()
					self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
			self.newlines = 0
			return self.buffer.read()
 
	def readline(self):
			while self.newlines == 0:
					b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
					self.newlines = b.count(b"\n") + (not b)
					ptr = self.buffer.tell()
					self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
			self.newlines -= 1
			return self.buffer.readline()
 
	def flush(self):
			if self.writable:
					os.write(self._fd, self.buffer.getvalue())
					self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
	def __init__(self, file):
			self.buffer = FastIO(file)
			self.flush = self.buffer.flush
			self.writable = self.buffer.writable
			self.write = lambda s: self.buffer.write(s.encode("ascii"))
			self.read = lambda: self.buffer.read().decode("ascii")
			self.readline = lambda: self.buffer.readline().decode("ascii")
 

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

from collections import defaultdict 
#from collections import deque
#from collections import OrderedDict
from math import gcd,log2,ceil
#import time
#import itertools
#import timeit
#import random
#from bisect import bisect_left as bl
#from bisect import bisect_right as br
#from bisect import insort_left as il
#from bisect import insort_right as ir
#from heapq import *
#mod=998244353
#mod=10**9+7

for _ in range(1):
    n,its=mi()
    a=li()
    s=sorted(set(a))
    s=list(s)
    if n*ceil(log2(len(s)))<=8*its:
        print(0)
        continue
    d=dd()
    for i in range(n):
        d[a[i]]+=1
    k=2**((8*its)//n)
    prfx=[0 for i in range(len(s)+1)]
    for i in range(1,len(prfx)):
        prfx[i]=prfx[i-1]+d[s[i-1]]
    mx=0
    for i in range(len(prfx)-1,k-1,-1):
        mx=max(prfx[i]-prfx[i-k],mx)
    print(n-mx)