def mul(n,m):  # compute n*m; n and m are nonnegative integers, with n < 2^32
    k = 32
    a = n
    b = m
    c = 0
    # invariant ab+c = nm and a < 2^k
    while k != 0:
        if a&1 == 1:
            # no need to do a = a - 1, because of later a=a>>1
            c = c + b
            # (a-1)b + (c+b) = ab -b + c + b = ab+c = nm
        a = a>>1
        b = b<<1
        # (a/2)2b+c = ab+c = nm
        k = k - 1
        # since the previous a was < the previous 2^k,
        #  a/2 < 2^(k-1)
        #  (a-1)/2 < 2^(k-1)
    # we know a < 2^0, so a=0, just like before
    return c