Which Squeeze to play for - never mind, there's only one possibility

Last night I had an interesting play problem in a slam. In 6NT, there are 11 top tricks. How would you plan on getting the 12^th?
 

			North ♠ AK8x ♥ AKx ♦ Ax ♣ Kxxx
			South ♠ Tx ♥ QJx ♦ KT8x ♣ AQJx
	South	West		North	East
	1♦	Pass		1♠	Pass
	1NT	Pass		6NT	Pass
	Pass	Pass

After the lead of the ♥T, I played a spade to my ten and LHO's Q to rectify the count and try to set up a squeeze. Back came a heart, and I then cashed the clubs. LHO discarded 2 low diamonds and RHO ditched a heart. Both followed to the 3^rd heart and to the next 2 rounds of spades, with the ♠J appearing from RHO on the last spade. Under the ♦A, LHO drops the J. Should I now play a diamond to the T or a diamond to the K at trick 12?

In this 2 card ending, I know one opponent has a good spade and there are 3 diamonds out. If RHO has the good spade, the contract is un-makeable because LHO would surely have ♦Qx left. So, we must assume that LHO has the good spade and RHO has 2 diamonds left. The odds are 2 to 1 in favor of him having Qx left rather than xx. With 3 cards out, there are 2 ways he can have Qx but only 1 way he can have xx.

So does this mean that finessing is the right play? Well, no. It's what I did and it took me awhile to figure out that this was clearly the wrong play. LHO wouldn't play the J under the A at trick 11 unless it was his last diamond or if he had only QJ left. And I had a whole count on the hand (except I didn't know for sure who had the remaining spade), so LHO had to have another diamond. Playing the J from Jx left would be giving me an opportunity to finesse that he could prevent by keeping the singleton J. Therefore, the only legitimate line of play was to play for the squeeze where LHO started with ♠Q9xx and ♦QJxx. I'm glad I finally figured that out. I knew it wasn't just a restricted choice available spaces problem but I couldn't figure out why until just now.