{ Zenit CK 2011/2012, uloha e)
  Riesenie by Zemco
     
  Dynamicke programovanie, DP[i,j] znamena, aby najvacsi
  otoceny stvorec ma dolny bod v mieste [i,j]?
  Pocitame postupne pre zvacsujuce sa i a zvacsujuce sa j.
  Cas O(R*C), pamat (R*C) dala by sa zlepsit.}
uses math;
var i,j,R,C: integer;
    M: array[1..1005,1..1005] of char;
    DP: array[1..1005,1..1005] of integer;
    best: integer;
{cely vstup mame zapamatany v poli o dva riadky a stlpce posunute
 pre zjednodusenie kodu. }
begin
  best := 0;
  readln(R,C);
  for i:=1 to R+2 do for j:=1 to C+2 do DP[i,j] := 0;
  for i:=1 to R+2 do for j:=1 to C+2 do M[i,j] := '.';
  for i:=1 to R do
  begin
    for j:=1 to C do read(M[i+2,j+2]);
    readln;
  end;
  for i:=1 to R+2 do for j:=1 to C+2 do
  begin
    if (M[i,j] = '.') then DP[i,j] := 0
    else DP[i,j] := min( min( DP[i-1,j-1], DP[i-1,j+1]), min( DP[i-2,j], DP[i-1,j]) ) + 1;
    best := max(best, DP[i,j]);
  end;
  if (best = 0) then writeln(0)
  else writeln( (2*best-1)*(2*best-1) - 2*(best*(best-1) ) );
end.