How would you build this cross?

[Zarxrax]

I got an interesting scramble today.
U' L D R U' B R D2 L F U' F' U2 R' U L2 F U2 R2 U2 F L2 D' F R'

The interesting thing about this scramble, is that 2 corner-edge pairs are already built right at the start. Now, of course, seeing this, I tried to exploit it and preserve the pairs while solving the cross. But, that slowed me down so badly that I ended up adding a good 10 seconds to my normal solve time.

If you come across a situation like this, would you try to keep these pairs, or would you just ignore them? If you would try to keep them, then how would you solve this particular cross?

[CubesOfTheWorld]

y' L' R y' U2 L' R' d R' U'. I still find f2l rather fast with this long cross. I always ignore the pairs, because when you have a pair, that pair will still be easy to pair when you have to pair them. lol

[iSpinz]

x2 R F' B U D2 L2 R'

[CubesOfTheWorld]

x2 R F' B U D2 L2 R'

lucky... you're good at the cross

[Zarxrax]

So basically, keeping those pairs are more trouble than they're worth, right?

[Meep]

x' (U D') L Lw2 z M' U' M' U x' Uw2

But I scramble with my cross color on top; I saw the two ready-made pairs on the opposite color. =( Either way though I only preserve pairs if it's easy to.

Edit:
Doing it with cross on bottom:

L' F' D L' F D

Again not preserving the pairs =(

[Sa967St]

xx-cross: (y M2 U2) (z2 F2 M' U2 M) (U' y Rw U2 R'w) (y2 L' R' F R)
kinda long... but it can be done fast

[nlCuber22]

L' F' D F L' D
Pairs aren't worth it imo, but (if you're scrambling yellow top red front) you have a very easy red green pair.

Spoiler:

Solution:
L' F' D F L' D
R' U R F' U F
U L' U2 L U' L' U L
U' R' U R U y R' U2 R U' R' U R
U R U' R' U2 R U' R'
F R U R' U' F'
U R' U2 R U2 R' F R U R' U' R' F' R2 U2

alternatively:
L' F' D F L' D
R' U R F' U F
U' y R' U' R
U y' R' U R L' U2 L F' L' F
U R' U' R
R' U2 R2 U R' U R U2 x' U' R' U
U R U R' U' R' F R2 U' R' U' R U R' F' U'

[ianini]

y' R L' U2 B' F U' R U
8 Moves

[amostay2004]

R F' D U' L' F D

The U' is to preserve 1 pair. That's pretty much what I would do. Preserving 2 pairs can be really hard in most scenarios. Also as you preserve that pair keep track on where the other pair went. They're often easy to pair up again + insert. In this case however, it's not