Problem: Diana loves playing with numbers. She's got n cards with positive integer numbers a_i written on them. She spends her free time multiplying the numbers on the cards. She picks a non-empty subset of the cards and multiplies all the numbers a_i written on them.

Diana is happy when the product of the numbers ends with her favorite digit d. Now she is curious what cards she should pick so that the product of the numbers on them is the largest possible and the last decimal digit of the product is d. Please, help her.

Input

The first line contains the integers n and d (1≤ n≤ 10^5, 0≤ d≤ 9). The second line contains n integers a_i (1≤ a_i≤ 1000). 

Output

On the first line, print the number of chosen cards k (1≤ k≤ n). On the next line, print the numbers written on the chosen cards in any order. 

If it is impossible to choose a subset of cards with the product that ends with the digit d, print the single line with -1.

Examples

Input


6 4
4 11 8 2 1 13


Output


5
1 2 4 11 13 


Input


3 1
2 4 6


Output


-1


Input


5 7
1 3 1 5 3


Output


-1


Input


6 3
8 9 4 17 11 5


Output


3
9 11 17 


Input


5 6
2 2 2 2 2


Output


4
2 2 2 2 

Note

In the first example, 1 × 2 × 4 × 11 × 13 = 1144, which is the largest product that ends with the digit 4. The same set of cards without the number 1 is also a valid answer, as well as a set of 8, 11, and 13 with or without 1 that also has the product of 1144.

In the second example, all the numbers on the cards are even and their product cannot end with an odd digit 1.

In the third example, the only possible products are 1, 3, 5, 9, 15, and 45, none of which end with the digit 7.

In the fourth example, 9 × 11 × 17 = 1683, which ends with the digit 3. 

In the fifth example, 2 × 2 × 2 × 2 = 16, which ends with the digit 6.

Solution (in PYTHON3):

Solution:
#------------------Important Modules------------------#
from sys import stdin,stdout
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import *
from random import *
input=stdin.readline
prin=stdout.write
from random import sample
from collections import Counter,deque
from fractions import *
from math import sqrt,ceil,log2,gcd
#dist=[0]*(n+1)
mod=10**9+7
"""
class DisjSet: 
    def __init__(self, n): 
        
        self.rank = [1] * n 
        self.parent = [i for i in range(n)] 
    def find(self, x): 
        if (self.parent[x] != x): 
              
             

            self.parent[x] = self.find(self.parent[x]) 
              
            
        return self.parent[x] 
  
  
    
    def union(self, x, y): 
          
         
        xset = self.find(x) 
        yset = self.find(y) 
  
        if xset == yset: 
            return
  
        if self.rank[xset] < self.rank[yset]: 
            self.parent[xset] = yset 
  
        elif self.rank[xset] > self.rank[yset]: 
            self.parent[yset] = xset 
  
        else: 
            self.parent[yset] = xset 
            self.rank[xset] = self.rank[xset] + 1
  

"""

def f(arr,i,j,d,dist):
    if i==j:
        return 
    nn=max(arr[i:j])
    for tl in range(i,j):
        if arr[tl]==nn:
            dist[tl]=d
            #print(tl,dist[tl])
            f(arr,i,tl,d+1,dist)
            f(arr,tl+1,j,d+1,dist)
    #return dist
            
        
def ps(n):
    cp=0;lk=0;arr={}
    #print(n)
    #cc=0;
    while n%2==0:
        n=n//2
        #arr.append(n);arr.append(2**(lk+1))
        lk+=1
    
    
    
        
    for ps in range(3,ceil(sqrt(n))+1,2):
        lk=0
        cc=0
        while n%ps==0:
            n=n//ps
            #cc=1
            #arr.append(n);arr.append(ps**(lk+1))
            lk+=1
        
            
            
    if n!=1:
        #lk+=1
        #arr[n]=1
        #ans*=n;
        lk+=1
        return False
        #return arr
    
    #print(arr)
    
    return True


#count=0
#dp=[[0 for i in range(m)] for j in range(n)]
#[int(x) for x in input().strip().split()]
def gcd(x, y): 
      
    while(y): 
        x, y = y, x % y 
      
    return x 
          
# Driver Code         
def factorials(n,r):
    #This calculates ncr mod 10**9+7
    slr=n;dpr=r
    qlr=1;qs=1
    mod=10**9+7
    
    for ip in range(n-r+1,n+1):
        qlr=(qlr*ip)%mod
    for ij in range(1,r+1):
        qs=(qs*ij)%mod
    #print(qlr,qs)
    ans=(qlr*modInverse(qs))%mod
    return ans

        
        
def modInverse(b):
    qr=10**9+7
    return pow(b, qr - 2,qr)

#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
  def wrapped_func(*args, **kwargs):
    if stack: return f(*args, **kwargs)
    to = f(*args, **kwargs)
    while True:
      if type(to) is GeneratorType:
        stack.append(to)
        to = next(to)
        continue
      stack.pop()
      if not stack: break
      to = stack[-1].send(to)
    return to
  return wrapped_func
def power(arr):
  listrep = arr
  subsets = []
  for i in range(2**len(listrep)):
    subset = []
    for k in range(len(listrep)):            
      if i & 1<<k:
        subset.append(listrep[k])
    subsets.append(subset)        
  return subsets
def pda(n) :
    list = [] 
    for i in range(1, int(sqrt(n) + 1)) :
         if (n % i == 0) :
            if (n // i == i) :
                list.append(i)
            else :
                list.append(n//i);list.append(i)
                 
    # The list will be printed in reverse    
    return list
def dis(xa,ya,xb,yb):
    return sqrt((xa-xb)**2+(ya-yb)**2)
#### END ITERATE RECURSION ####
 
#===============================================================================================
#----------Input functions--------------------#
def ii():
    return int(input())
def ilist():
    return [int(x) for x in input().strip().split()]
def islist():
    return list(map(str,input().split().rstrip()))
def inp():
    return input().strip()
def google(test,ans):
    return "Case #"+str(test)+": "+str(ans);
def overlap(x1,y1,x2,y2):
    if x2>y1:
        return y1-x2
    if y1>y2:
        return y2-x2
    return y1-x2;


###-------------------------CODE STARTS HERE--------------------------------###########
def pal(s):
    k=len(s)
    n=len(s)//2
    for i in range(n):
        if s[i]==s[k-1-i]:
            continue
        else:
            return 0
    return 1


            
#########################################################################################
#t=int(input())
t=1
for p in range(t):
    n,kll=ilist()
    arr=[-1]
    arr.extend(ilist())
    dp=[[0 for i in range(10)] for j in range(n+1)]
    pointer=[[[-1 for k in range(3)] for i in range(10)] for j in range(n+1)] 
    pp=[]
    for i in range(1,n+1):
        for j in range(10):
            #dp[i][j]=dp[i-1][j]
            last=arr[i]%10
            
            for k in range(10):
                #print(k*last,j)
                #fs=dp[i-1][k]%10
                fs=1 if dp[i-1][k]==0 else k
                if (fs*last)%10==j:
                    if dp[i][j]<dp[i-1][k]+log2(arr[i]+0.01):
                        
                        pointer[i][j][0]=0
                        pointer[i][j][1]=i-1
                        pointer[i][j][2]=k
                        dp[i][j]=max(dp[i-1][k]+log2(arr[i]+0.01),dp[i][j])
                        #print("dp[i-1][k] ",dp[i-1][k],"dp[i][j] ",dp[i][j],"j ",j,"i ",i,"k ",k)
            if dp[i][j]<dp[i-1][j]:
                pointer[i][j][0]=1
                pointer[i][j][1]=i-1
                pointer[i][j][2]=j
                dp[i][j]=max(dp[i-1][j],dp[i][j])
                        
            
    ans=dp[n][kll]
    if ans==0:
        print(-1)
        continue
    #print(ans)
    ap=n;jp=kll
    count=0
    while True:
        up=ap
        
        if pointer[ap][jp][0]!=0:
            
            ap=pointer[up][jp][1]
            jp=pointer[up][jp][2]
        elif pointer[ap][jp][0]==0:
            pp.append(arr[ap])
            count+=1
            ap=pointer[up][jp][1]
            jp=pointer[up][jp][2]
        
        if ap<=0:
            break
        
    print(count)
    print(*pp)
            
#byevye