module Mult(input wire clock, input wire reset, input wire[31:0) InA, input wire[31:0) InB,
            input wire MultControl,
            output reg[31:0} Hi, output reg[31:0} Lo, output reg MultExit);
            --Baseado no algoritmo de booth https://en.wikipedia.org/wiki/Booth%27s_multiplication_algorithm#A_typical_implementation

    integer i = -1;
    reg [64:0] A;  --Formato Geral [x + y + 1] -> 32 + 32 + 1 = 65
    reg [64:0] S;
    reg [64:0] P;
    reg [31:0] TwoComp;

    always @(posedge clk) begin
        if(reset) begin
            Hi = 32'b0; --ZERANDO $HI
            Lo = 32'b0; --ZERANDO $LO
            A = 65'b0; --Zerando A
            S = 65'b0; --Zerando S
            P = 65'b0; --Zerando P
            i = -1; --Zerando o iterador
        end

        if(MultControl == 1'b1) begin  --Sinal de comeco
            i = 0;   --for( -> int i = 0 <- ; i < 32; i++)
            A = {InA, 33'b0}; --Fill the most significant (leftmost) bits with the value of m.
            --Fill the remaining (y + 1) bits with zeros.
            TwoComp = (~a + 1'b1); --Notacao de Complemento a 2
            S = {TwoComp, 33'b0}; --Fill the most significant bits with the value of (−m) in
            --two's complement notation. Fill the remaining (y + 1) bits with zeros.
            P = {32'b0, b, 1'b0}; --Fill the most significant x bits with zeros. To the right of
            --this, append the value of r. Fill the least significant (rightmost) bit with a zero.
            MultExit = 1'b0; --Sinal de Saida = 0
        end

        case (P[1:0])
        --Determine the two least significant (rightmost) bits of P.
        --  If they are 01, find the value of P + A. Ignore any overflow.
        2'b01: P = P + A;
        --  If they are 10, find the value of P + S. Ignore any overflow.
        2'b10: P = P + S;
        --  If they are 00, do nothing. Use P directly in the next step.
        --  If they are 11, do nothing. Use P directly in the next step.
        endcase
        P = P >> 1; --Arithmetically shift the value obtained in
        --the last step by a single place to the right. Let P now equal this new value.

        if(P[63] == 1'b1) begin --Caso do Numero Negativo
            P[64] = 1'b1;
        end

        if(i < 32) begin --Iteracao do i  for(int i = 0; i < 32; --> i++ <--)
            i = i + 1;
        end

        if(i == 32) begin --Algum momento i sera 32, ai os registradores serao setados e os valores resetados
            Hi = P[64:33]; --Assign do Hi
            Lo = P[32:1]; --Assign do Lo
            MultEnd = 1'b1; --sinal de saida = 1

            i = -1; --RESETING i
            A = 65'b0; --RESETING A
            S = 65'b0; --RESETING S
            P = 65'b0;  --RESETING P
        end
    end

endmodule