Количество пар кр и не кр 11 оптимизированное с логикой, Pascal

Количество пар кр и не кр 11 оптимизированное с логикой

//fpc 3.0.0

program HelloWorld;

const 
    delta = 1;

var
    n, i, s, s11, sn, x : integer;
    m : array [0..delta] of integer;

begin
    s := 0; s11 := 0; sn := 0;
    readln (n);
    for i := 0 to delta do begin readln (x); m[i] := x end;
    
    for i := delta + 1 to n - 1 do
    begin
    
        readln (x);
        if (m[0] mod 11 = 0) and (x mod 11 = 0) then
        begin
            s11 := s11 + 1;
            s := s + s11 + sn
        end
        else if (m[0] mod 11 <> 0) and (x mod 11 = 0) then
        begin
            sn := sn + 1;
            s := s + s11 + sn
        end
        else if (m[0] mod 11 = 0) and (x mod 11 <> 0) then
        begin
            s11 := s11 + 1;
            s := s + s11
        end
        else if (m[0] mod 11 <> 0) and (x mod 11 <> 0) then
        begin
            sn := sn + 1;
            s := s + s11
        end;
        
        
        {      
        if m[0] mod 11 = 0 then s11 := s11 + 1 else sn := sn + 1;
        readln (x);
        if x mod 11 = 0 then s := s + s11 + sn else s := s + s11;
        }
        m[0] := m[1]; m[1] := x;
    end;
    writeln(s);
end.


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