nt.number theory - The large sieve for primes

Let $Lambda(n)$ be the von Mangoldt function, i.e., $Lambda(n) = log p$ for $n$ a prime power $p^k$ and $Lambda(n) = 0$ for all $n$ that not prime powers. Let

$$S(alpha) = sum_{n leq N} Lambda(n) e(alpha n).$$

Now, using, say, Lemma 7.15 in Iwaniec-Kowalski (or the same result in Montgomery), we get

$$sum_{q leq q_0} sum_{a pmod{q}: gcd(a,q)=1} lvert S(a/q)rvert^2
leq frac{(N + Q^2) N log N}{sum_{substack{qleq Q text{ squarefree} \ gcd(q,P(q_0))=1}} phi(q)^{-1}},$$

where $Q$ is arbitrary and $P(z):=prod_{p leq z} p$.

In practice, we would choose $Q$ slightly smaller than $sqrt{N}$, and obtain

$$sum_{q leq q_0} sum_{substack{a pmod{q} \ gcd(a,q)=1}} lvert S(a/q) rvert^2 leq (1+epsilon) 2 e^gamma N^2 log q_0,$$

where gamma is Euler's constant $0.577cdots$ and $epsilon$ is very small.

Now, the 2 in the bound $leq (1+epsilon) 2 e^gamma N^2$ is due to the parity problem,
and thus should be next to impossible to remove (except for very small $q_0$).
However, the factor of $e^gamma$ clearly has no right to exist. The true asymptotic should be simply $N^2 log q_0$.

Can we remove that nasty $e^gamma$? That is, can you prove a bound of type

$$sum_{q leq q_0} sum_{substack{a pmod{q} \ gcd(a,q)=1}} lvert S(a/q)rvert^2 leq (1+epsilon) 2 N^2 log q_0 ?$$

Harald