{完整版：無解，多解，唯一解}
const max=100;
var
  a:array[1..max,1..max+1] of real;
  x:array[1..max] of real;
  n,pre:integer;
  procedure init;
    var i,j:integer;
    begin
      assign(input,'a4.in');reset(input);
      readln(n);
      for i:=1 to n do
        for j:=1 to n+1 do read(a[i,j]);
      close(input);
    end;
  procedure change(i,pre:integer);{一次消元：第i行，第pre列開始}
    var j,k:integer;
        tem:real;
    begin
      for j:=i+1 to n do
            begin
              tem:=a[j,pre]/a[i,pre];{乘數(shù)}
              for k:=pre to n+1 do
              a[j,k]:=a[j,k]-a[i,k]*tem;
            end;
    end;
  function find(i:integer):integer;{返回第i行，第一個不為0的數(shù)所在的列，包括交換調(diào)整后}
    var row,col,j,k:integer;
        tem:real;
    begin
          row:=i;
          for col:=pre+1{1} to n+1 do {從1開始較慢，從上次的pre下一列開始即可}
            begin
              for j:=i+1 to n do {找一個絕對值大的數(shù)}
                if abs(a[j,col])>abs(a[row,col]) then row:=j;
              if abs(a[row,col])>0 then {找到非0數(shù)}
                 begin
                   if row<>i then {交換行}
                      for k:=col to n+1 do
                        begin  tem:=a[i,k];a[i,k]:=a[row,k];a[row,k]:=tem;end;
                   find:=col;
                   exit;
                 end;
            end;
          find:=0;{已無非0數(shù)}
    end;
  procedure xiaoyuan;
    var i,j,k,row,col,found:integer;
        tem:real;
    begin
      pre:=0;
      for i:=1 to n-1 do
        begin
          pre:=find(i);
          if pre=0 then exit;{已消元結(jié)束，后面全為0}
          change(i,pre);
        end;
    end;
  procedure juzhen;{輸出最后消元后的矩陣，以供檢驗}
    var i,j:integer;
    begin
      for i:=1 to n do begin
        for j:=1 to n+1 do write(a[i,j]:0:0,' ');writeln;end;
    end;
  procedure print(k:integer);
    var i,j:integer;
    begin
      if k<>1 then begin writeln(k);exit;end;
      x[n]:=a[n,n+1]/a[n,n];
      for i:=n-1 downto 1 do
        begin
          for j:=i+1 to n do
            a[i,n+1]:=a[i,n+1]-a[i,j]*x[j];
          x[i]:=a[i,n+1]/a[i,i];
        end;
      for i:=1 to n do
        writeln('x',i,'=',x[i]:0:2);
    end;
  procedure check;
    var i:integer;
    begin
      i:=n;
      while abs(a[i,n])<1e-10 do dec(i);{writeln(i); }
      if i=n then print(1);{有唯一的解}
      if (i<n)and(a[i+1,n+1]=0) then print(0);{無窮多組實數(shù)解}
      if (i<n)and(a[i+1,n+1]<>0) then print(-1);{無解}
    end;
BEGIN
  init;
  xiaoyuan;
  juzhen;
  check;
END.