Определить среднюю из трех заданных дату рождения

uses crt;
type date=record
          nm:string;
          gd,ms,dn:integer;
          end;
procedure Vvod(var d:date;s:string);
var k:byte;
begin
writeln(s);
repeat
write(' Год (1990-2010):');readln(d.gd);
until (d.gd>=1990)and(d.gd<=2010);
repeat
write('  Месяц (1-12): ');readln(d.ms);
until d.ms in [1..12];
repeat
k:=0;
write('  День: ');readln(d.dn);
if ((d.ms in [1,3,5,7,8,10,12])and not(d.dn in [1..31]))
or ((d.ms in [4,6,9,11])and not(d.dn in [1..30]))
or ((d.ms=2)and(d.gd mod 4=0)and not(d.dn in [1..29]))
or ((d.ms=2)and(d.gd mod 4<>0)and not(d.dn in [1..28]))then k:=1;
until k=0;
end;
var dt:array[1..3]of date;
    x:date;
    i,j:byte;
begin
dt[1].nm:='Дмитрий';
Vvod(dt[1],dt[1].nm);
dt[2].nm:='Иван';
Vvod(dt[2],dt[2].nm);
dt[3].nm:='Олег';
Vvod(dt[3],dt[3].nm);
for i:=1 to 2 do
for j:=i+1 to 3 do
if (dt[i].gd<dt[j].gd)or((dt[i].gd=dt[j].gd)and(dt[i].ms<dt[j].ms))
or((dt[i].gd=dt[j].gd)and(dt[i].ms=dt[j].ms)and(dt[i].dn<dt[j].dn)) then
 begin
  x:=dt[i];
  dt[i]:=dt[j];
  dt[j]:=x
 end;
write('Средний ',dt[2].nm);
end.