Barban-Davenport-Halberstam without von Mangoldt weights

The Barban-Davenport-Halberstam theorem gives a bound for the average (in L_2 norm) difference between $\sum_{n\leq N: n\equiv a \mod q} \Lambda(n)$ and $N/\phi(q)$. It is obvious that a similar result should hold for the difference between $\sum_{p\leq N: p\equiv a \mod q} 1$ (where $p$ ranges only across primes) and $\pi(N)/\phi(q)$. Does anybody know where in the literature a statement in that form can be found (so that it can be quoted without any further ado - the alternative is to spend some space in its derivation)?

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I don't know of any such cut and dried reference but I find dealing with von-Mangoldt easier than dealing with primes. –  Idoneal Oct 26 '10 at 4:29
Yes, that's why the von Mangoldt function was ever defined. At the same time, we sometimes have to deal with primes! –  H A Helfgott Oct 26 '10 at 8:59
Well said Mukherjee! –  Idoneal Oct 26 '10 at 10:36
This is obviously a cultural reference I am missing. At any rate, can we get back to the question? –  H A Helfgott Oct 26 '10 at 11:58
Have you seen Theorem 17.5 of Iwaniec-Kowalski? I think that is all you need. –  Idoneal Oct 26 '10 at 13:33
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Theorem 17.5 of Iwaniec-Kowalski seems to do the job.

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For the benefit of anyone else reading - what exactly is Iwaniec-Kowalski? –  Gerry Myerson Oct 26 '10 at 23:07
It is the modern bible of analytic number theory. books.google.com/… –  Idoneal Oct 27 '10 at 4:25
Thanks. To save others the trouble of clicking through, it's Henryk Iwaniec and Emmanuel Kowalski, Analytic Number Theory, American Mathematical Society Colloquium Publications Volume 53. –  Gerry Myerson Oct 27 '10 at 5:22