"""

https://yukicoder.me/problems/no/1189

各桁1回づつしか使えない
各桁が重ならないような取り方の数を数える
Aも実は1000通りぐらいしかない
あとはdp
1000*1000ぐらい

"""

from sys import stdin
import sys


def modfac(n, MOD):
 
    f = 1
    factorials = [1]
    for m in range(1, n + 1):
        f *= m
        f %= MOD
        factorials.append(f)
    inv = pow(f, MOD - 2, MOD)
    invs = [1] * (n + 1)
    invs[n] = inv
    for m in range(n, 1, -1):
        inv *= m
        inv %= MOD
        invs[m - 1] = inv
    return factorials, invs


def modnCr(n,r,mod,fac,inv): #上で求めたfacとinvsを引数に入れるべし(上の関数で与えたnが計算できる最大のnになる)

    return fac[n] * inv[n-r] * inv[r] % mod

def cdig(n):
    ret = 0
    for i in range(10):
        if (2**i) & n > 0:
            ret += 1
    return ret

mod = 998244353
N,K = map(int,stdin.readline().split())
A = list(map(int,stdin.readline().split()))
nums = [0] * 1024

if K >= 10:
    print (0)
    sys.exit()

facs,invs = modfac(K+10,mod)

for i in A:
    nums[i] += 1

dp = [ [0]*(K+1) for i in range(1024) ]
dp[0][0] = 1


tmps = [ (cdig(i),i) for i in range(1024) ]
tmps.sort()

for dnum,x in tmps:

    for i in range(1024):
        for nk in range(K):
            if x & i == 0:
                dp[x | i][nk+1] += dp[x][nk] * nums[i]
                dp[x | i][nk+1] %= mod

ans = 0
for i in range(1024):
    ans += dp[i][-1]
    ans %= mod
print ((ans * invs[K]) % mod)