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 sys
  import math
  import heapq
  import bisect
  from collections import Counter
  from collections import defaultdict
  from io import BytesIO, IOBase
  from itertools import permutations
  import string
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          import os
          self.os = os
          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
          self.BUFSIZE = 8192
  
      def read(self):
          while True:
              b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.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 = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.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:
              self.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")
  
  
  def get_int():
      return int(input())
  
  
  def get_ints():
      return list(map(int, input().split(' ')))
  
  
  def get_int_grid(n):
      return [get_ints() for _ in range(n)]
  
  
  def get_str():
      return input().strip()
  
  
  def get_strs():
      return get_str().split(' ')
  
  
  def flat_list(arr):
      return [item for subarr in arr for item in subarr]
  
  
  def yes_no(b):
      if b:
          return "YES"
      else:
          return "NO"
  
  
  def binary_search(good, left, right, delta=1, right_true=False):
      """
      Performs binary search
      ----------
      Parameters
      ----------
      :param good: Function used to perform the binary search
      :param left: Starting value of left limit
      :param right: Starting value of the right limit
      :param delta: Margin of error, defaults value of 1 for integer binary search
      :param right_true: Boolean, for whether the right limit is the true invariant
      :return: Returns the most extremal value interval [left, right] which is good function evaluates to True,
              alternatively returns False if no such value found
      """
  
      limits = [left, right]
      while limits[1] - limits[0] > delta:
          if delta == 1:
              mid = sum(limits) // 2
          else:
              mid = sum(limits) / 2
          if good(mid):
              limits[int(right_true)] = mid
          else:
              limits[int(~right_true)] = mid
      if good(limits[int(right_true)]):
          return limits[int(right_true)]
      else:
          return False
  
  
  def prefix_sums(a):
      p = [0]
      for x in a:
          p.append(p[-1] + x)
      return p
  
  
  def solve_a():
      l, r = get_ints()
      if l <= r // 2 + 1:
          return (r - 1) // 2
      else:
          return r - l
  
  
  def substring(s, t):
      n, m = len(s), len(t)
      i = 0
      j = 0
      while i < n and j < m:
          if s[i] == t[j]:
              i += 1
          j += 1
      return i == n
  
  
  
  def isprime(n):
      if n == 1:
          return False
      i = 2
      while i * i <= n:
          if n % i == 0:
              return False
          i += 1
      return True
  
  
  def solve_b():
      n = get_int()
      s = list(get_str())
      for i in range(1, 500):
          if not isprime(i) and substring(str(i), s):
              print(len(str(i)))
              print(i)
              return
  
  
  
  
  
  
  
  
  t = get_int()
  for _ in range(t):
      solve_b()
  