Adding 3, and still no change? Strange!

A number, when divided by 11, gives a remainder KK. If the digit 33 is placed at the end of the number, the remainder when divided by 11 still remains KK.

Find the remainder when K(K1)(K2)321\displaystyle K^{(K-1)^{(K-2)^{\dots^{3^{2^1}}}}} is divided by 1111.

Details and assumptions:-

The number is in base 10.

Placing the digit 5 at the end of 123123 makes the number into 12351235.

If KK was 55, you had to find the remainder when 543215^{4^{3^{2^1}}} is divided by 1111.

This is a part of the set 11≡ awesome (mod remainders)


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