捕获1位数字，但也带有其他条件+正则表达式

如何0从行K'0
中捕获要捕获的其他所有内容。

这是我的正则表达式示例

这是我的正则表达式
K'(?P<name1>81|61|64|44|86|678|41|49|33|685|1(?:33|45)?|\d{1,3})?\d+

K'0     <<<----adding the ? here |\d{1,3})?\d+ as want to pick up if there is only 
K'93        <<<--- 1 number 2 number or 3 numbers (ie. K'0, K'93, K'935 )
K'935
K'8134567
K'81345678
K'6134516789
K'61345678
K'643456
K'646345678
K'1234567890
K'12345678901
K'1454567890    <<<--- want 145 returned and not 1 
K'13345678901   <<<--- want 133 returned and not 1 
K'3214567890123
K'32134567890123
K'3654567890123
K'8934567890123
K'6554567890123


I am interested in the digits after K'
I am looking to do this using regex but not sure if it can be done. What I want is:
if the number starts with 81 return 81 
if the number starts with 61 return 61 
...
if the number starts with something i am not interested in return other(or its first digits of 1-3)

The above criteria works:

but what I also want is:
if the fist digit is 1 then return 1 BUT 
if the fist digit is 1 and the 2nd and 3rd digit are 45 return 145 and don't return just 1  
if the fist digit is 1 and the 2nd and 3rd digit are 33 return 133 and don't return just 1

I presume I have to put something inside this part of the regex |(1)\d+|

Questions:
Does regex sort the data first?
Is the order of the regex search important to how it is implemented? i deally I do not want this.