The following cryptarithm is a multiplication problem that can be solved by substituting digits from a specified set of N digits into the positions marked with *. If the set of prime digits {2,3,5,7} is selected, the cryptarithm is called a PRIME CRYPTARITHM.

      * * *
   x    * *
    -------
      * * *
    * * *
    -------
    * * * *

Write a program that will find all solutions to the cryptarithm above for any subset of digits from the set {1,2,3,4,5,6,7,8,9}.

PROGRAM NAME: crypt1

INPUT FORMAT

SAMPLE INPUT (file crypt1.in)

5
2 3 4 6 8

OUTPUT FORMAT

A single line with the total number of unique solutions. Here is the single solution for the sample input:

      2 2 2
    x   2 2
     ------
      4 4 4
    4 4 4
  ---------
    4 8 8 4

SAMPLE OUTPUT (file crypt1.out)

1

{
ID:style71
PROG:crypt1
LANG:PASCAL
}
program crypt1;
var  p1,p2,p3,p4,p5,n,i:integer;
       sum:longint;
       f1,f2:text;
       num:array [1..9] of integer;
function ifin(k:integer):boolean;
  var j:integer;
  begin
      for j:=1 to n do
        if num[j]=k
          then
            begin
              ifin:=true;
              exit;
            end;
      ifin:=false;
  end;
function pan(x:integer):boolean;
  var l:integer;
  begin
      repeat
        begin
          l:=x mod 10;
          if not(ifin(l))
            then
              begin
                pan:=false;
                exit;
              end;
          x:=x div 10;
        end;
      until x<1;
      pan:=true;
  end;

function try:boolean;
  var a,b,s1,s2,s:integer;
  begin
      a:=num[p1]*100+num[p2]*10+num[p3];
      b:=num[p4]*10+num[p5];
      s1:=a*num[p4];
      s2:=a*num[p5];
      if (s1>=1000) or (s2>=1000)
        then
          begin
            try:=false;
            exit;
          end;
      if (not(pan(s1)))
        then
          begin
            try:=false;
            exit;
          end;
      if (not(pan(s2)))
        then
          begin
            try:=false;
            exit;
          end;
      s:=s1*10+s2;
      if (s>=10000) or (not(pan(s)))
        then
          begin
            try:=false;
            exit;
          end;
      try:=true;
  end;
begin
  assign(f1,'crypt1.in');
  assign(f2,'crypt1.out');
  reset(f1);
  rewrite(f2);
  readln(f1,n);
  sum:=0;
  for i:=1 to n do
      read(f1,num[i]);
  for p1:=1 to n do
      for p4:=1 to n do
        if (num[p1]*num[p4]<10)
          then
            for p5:=1 to n do
              if (num[p1]*num[p5]<10)
                then
                  for p2:=1 to n do
                    for p3:=1 to n do
                      if try
                        then
                          inc(sum);
  writeln(f2,sum);
  close(f1);
  close(f2);
end.