energy of two colliding mosquitos

Today is the day or maybe its tomorrow that they switch on the Large Hadron Collider. There was an amusing quote today I decided to check out.

"Each collision of a pair of protons in the LHC will release an amount of energy comparable to that of two colliding mosquitoes, so any black hole produced would be much smaller than those known to astrophysicists." Whether or not we will see a black hole or not, I will leave open for debate but I will do my best to figure the rest.

First the easy part, what is the energy of two colliding mosquitoes?
Mass of a mosquito 2E-6kg (two of them.... 4E-6kg) http://en.wikipedia.org/wiki/Orders_of_magnitude_(mass) (light little buggers)
Speed of a mosquito 2km/h = .5555m/s (assume they are heading towards each other 1.1111m/s (http://wiki.answers.com/Q/What_is_the_flight_speed_of_a_mosquito) I am not sure that this figure is laden or unladen....

So we can get a kinetic energy of the system at m*v*v/2 (4E-6)*1.1111*1.1111/2 = 2.46909E-06 Joules

Now the harder part.....
Mass of a proton 1.67262158E-27kg
Speed of light...299792458 m/s

I think there are two ways to measure the energy of a system of two colliding relativistic particles. I think E=mc2 is the right one vs kinetic energy (which is close enough). I do not believe that the 'relative' velocities need to be taken into account either due to relativistic effects (not that it will make any difference). so the energy of the system is (2m)c*c which yields

3.01E-10Joules

Apparently, they either used huge/engorged or really fast mosquitoes or my math is off, but the energy of two colliding mosquitoes is 10000 times greater than that of collision of two magnificently accelerated protons.

Apparently the world will be safe for another day. At least until the heavy ion (Pb) collision experiments start....