def mul(n,m):  # compute n*m; n and m are nonnegative integers, with n < 2^32
    k = 32
    # new variables:
    #  d = b/2^(32-k), so b = d*2^(32-k)
    #  e = c*2^k + a
    #a = n
    #b = m
    #c = 0
    d = m   # because d = b/2^(32-k) = m/2^(32-32) = m
    e = n   # because e = c*2^k + a = 0*2^32 + n = n
    # invariant ab+c = nm and a < 2^k
    while k != 0:
        if e&1 == 1:
            # no need to do a = a - 1, because of later a=a>>1
            # c = c + b
            # e' = c'*2^k + a
            #    = (c+b)*2^k + a
            #    = c*2^k + b*2^k + a
            #    = (c*2^k + a) + b*2^k
            #    = e + b*2^k
            #    = e + (d*2^(32-k))*2^k
            #    = e + d*2^32
            e = e + (d<<32)
            # (a-1)b + (c+b) = ab -b + c + b = ab+c = nm
        #a = a>>1
        #b = b<<1
        # d doesn't need updating, because b doubled and k decreases by 1
        e = e>>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 e