f(i) = sum{f(k1)+1, f(k2)+1, ...}(a[i]>max{a[k1],a[k2],a[k3],..} && ki < n)
k1=i+1;
k2=k1+f(k1)+1
k3=k2+f(k2)+1

res = sum{f(0),f(1),...f(n-1)}