description: During the hypnosis session, Nicholas suddenly remembered a positive integer n, which doesn't contain zeros in decimal notation. 
  
  Soon, when he returned home, he got curious: what is the maximum number of digits that can be removed from the number so that the number becomes not prime, that is, either composite or equal to one?
  
  For some numbers doing so is impossible: for example, for number 53 it's impossible to delete some of its digits to obtain a not prime integer. However, for all n in the test cases of this problem, it's guaranteed that it's possible to delete some of their digits to obtain a not prime number.
  
  Note that you cannot remove all the digits from the number.
  
  A prime number is a number that has no divisors except one and itself. A composite is a number that has more than two divisors. 1 is neither a prime nor a composite number.
  
  Input
  
  Each test contains multiple test cases.
  
  The first line contains one positive integer t (1 ≤ t ≤ 10^3), denoting the number of test cases. Description of the test cases follows.
  
  The first line of each test case contains one positive integer k (1 ≤ k ≤ 50) — the number of digits in the number.
  
  The second line of each test case contains a positive integer n, which doesn't contain zeros in decimal notation (10^{k-1} ≤ n < 10^{k}). It is guaranteed that it is always possible to remove less than k digits to make the number not prime.
  
  It is guaranteed that the sum of k over all test cases does not exceed 10^4.
  
  Output
  
  For every test case, print two numbers in two lines. In the first line print the number of digits, that you have left in the number. In the second line print the digits left after all delitions. 
  
  If there are multiple solutions, print any.
  
  Example
  
  Input
  
  
  7
  3
  237
  5
  44444
  3
  221
  2
  35
  3
  773
  1
  4
  30
  626221626221626221626221626221
  
  
  Output
  
  
  2
  27
  1
  4
  1
  1
  2
  35
  2
  77
  1
  4
  1
  6
  
  Note
  
  In the first test case, you can't delete 2 digits from the number 237, as all the numbers 2, 3, and 7 are prime. However, you can delete 1 digit, obtaining a number 27 = 3^3.
  
  In the second test case, you can delete all digits except one, as 4 = 2^2 is a composite number.
solution: import itertools
  from sys import maxsize, stdout, stdin,stderr
  mod = int(1e9+7)
  import sys
  def I(): return int(stdin.readline())
  def lint(): return [int(x) for x in stdin.readline().split()]
  def S(): return list(map(str,input().strip()))
  def grid(r, c): return [lint() for i in range(r)]
  from collections import defaultdict, Counter, deque
  import math
  import heapq
  from heapq import heappop , heappush
  import bisect
  from math import inf
  from itertools import groupby
  from itertools import permutations as comb
  def gcd(a,b): 
      while b:
          a %= b
          tmp = a
          a = b
          b = tmp
      
      return a
   
  def lcm(a,b): 
      return a  // gcd(a, b) * b
   
  def check_prime(n):
      for i in range(2, int(n ** (1 / 2)) + 1):
          if not n % i:
              return False
      return True
  
  def nCr(n, r):
   
      return (fact(n) // (fact(r)
                  * fact(n - r)))
   
  # Returns factorial of n
  def fact(n):
   
      res = 1
       
      for i in range(2, n+1):
          res = res * i
           
      return res
  def primefactors(n):
      num=0
      
      while n % 2 == 0:
          num+=1
          n = n / 2
      
      for i in range(3,int(math.sqrt(n))+1,2):
           
      
          while n % i== 0:
              num+=1
              n = n // i
                 
      
      if n > 2:
          num+=1
      return num
  '''
  def iter_ds(src):
      store=[src]
      while len(store):
          tmp=store.pop()
          if not vis[tmp]:
              vis[tmp]=True
              for j in ar[tmp]:
                  store.append(j)
  '''
  def ask(a):
      print('? {}'.format(a),flush=True)
      n=I()
      
      return n
  def linear_sieve(n):
      is_composite=[False]*n
      prime=[]
      for i in range(2,n):
          if not is_composite[i]:
              prime.append(i)
          for j in prime:
              is_composite[i*j]=True
              if i%prime==0:
                  break
      return prime
   
  def dfs(i,p,d):
      
      a,tmp=0,0
      for j in d[i]:
          if j!=p:
              a+=1
              tmp+=dfs(j,i)
      
      if a==0:
          return 0
      
      return tmp/a + 1 
  def primeFactors(n):
      l=[]
      
      while n % 2 == 0:
          l.append(2)
          n = n // 2
      for i in range(3,int(math.sqrt(n))+1,2):
           
          
          while n % i== 0:
              l.append(i)
              n = n // i
  
      if n > 2:
          l.append(n)
      return l
  
  
      
  
  #  Sieve
  d=[]
  primes=[]
  prim=[0]*(10**5+1)
  def sieve(n):
      
      for i in range(n):
          d.append(i)
  
      for i in range(2,n):
  
          if d[i]==i:
              prim[i]=1
              primes.append(i)
          j=0
          while j<len(primes) and primes[j]<=d[i] and primes[j]*i<n:
          
              d[i * primes[j]] = primes[j]
              j+=1
  
          
  def primeFactors(n):
      factors=[]
      while n!=1:
          factors.append(d[n])
          n//=d[n]
      
      return factors
  def maxPrefix(s, t) :
      count = 0
   
      # Iterating string T.
      for i in range(0,len(t)) :
           
          # If end of string S.
          if (count == len(s)) :
              break
   
          # If character match,
          # increment counter.
          if (t[i] == s[count]) :
              count = count + 1
               
   
      return count
  sieve(10**5)
  t=I()
  for _ in range(t):
      n=I()
      a=input()
      for i in range(1,min(int(a),10**5)+1):
          if prim[i]==0:
              tmp=maxPrefix(str(i),a)
              if tmp==len(str(i)):
                  print(tmp)
                  print(i)
                  break
  
  
  
  
  
      
      