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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