var P: array[1..100]of longint;
    x,i,j,N: longint;
    odpoved: int64;
{Zenit CK 2011/12, priklad C.
 
Vyuzijeme, ze hodnoty na vstupe su z rozsahu 1 az 100. 
P je pocet cisel danej hodnoty, ktore sme doteraz nacitali.
Ked nacitame cislo 'x', to tvori inverziu so vsetkymi cislami,
ktore sa doteraz vyskytli a boli vacsie.
}
begin
  readln(N);
  for i:=1 to 100 do P[i] := 0;
  odpoved := 0;
  for i:=1 to N do
  begin
    read(x);
    for j:=x+1 to 100 do odpoved := odpoved + P[j];
    inc(P[x]);
  end;
  writeln(odpoved);
end.
 

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