约翰的N(1≤N≤105)头奶牛站成一排，奶牛i的身高是Hi(l≤Hi≤1,000,000)．现在，每只奶牛都在向左看齐．对于奶牛i，如果奶牛j满足i<j且Hi<Hj，我们可以说奶牛i可以仰望奶牛j．    求出每只奶牛离她最近的仰望对象．

 

    第1行输入N，之后每行输入一个身高．

 

    共N行，按顺序每行输出一只奶牛的最近仰望对象．如果没有仰望对象，输出0．

 var

    i,j,k,l,m,n:longint;

    a,b:array[..] of longint;

 function max(x,y:longint):longint;inline;

          begin

               if x>y then max:=x else max:=y;

          end;

 function min(x,y:longint):longint;inline;

          begin

               if x<y then min:=x else min:=y;

          end;

 procedure built(z,x,y:longint);inline;

           begin

                if x=y then

                   begin

                        read(a[z]);

                        b[x]:=a[z];

                   end

                else

                    begin

                         built(z*,x,(x+y) div );

                         built(z*+,(x+y) div +,y);

                         a[z]:=max(a[z*],a[z*+]);

                    end;

           end;

 function approach(z,x,y,l,r,t:longint):longint;inline;

          var a1:longint;

          begin

               if l>r then exit();

               if a[z]<=t then exit();

               if x=y then

                  begin

                       if a[z]>t then exit(x);

                  end;

               a1:=approach(z*,x,(x+y) div ,l,min(r,(x+y) div ),t);

               if a1<> then exit(a1);

               exit(approach(z*+,(x+y) div +,y,max((x+y) div +,l),r,t));

          end;

 begin

      readln(n);

      built(,,n);

      for i:= to n do

          writeln(approach(,,n,i+,n,b[i]));

 end.