--- Part Two ---

The final step in breaking the XMAS encryption relies on the invalid number you just found: you must
[1m[37mfind a contiguous set of at least two numbers[0m in your list which sum to the invalid
number from step 1.

Again consider the above example:

35 20 [1m[37m15[0m [1m[37m25[0m [1m[37m47[0m [1m[37m40[0m 62 55 65 95 102 117 150 182
127 219 299 277 309 576

In this list, adding up all of the numbers from 15 through 40 produces the invalid number from step
1, 127. (Of course, the contiguous set of numbers in your actual list might be much longer.)

To find the [1m[37mencryption weakness[0m, add together the [1m[37msmallest[0m and
[1m[37mlargest[0m number in this contiguous range; in this example, these are 15 and 47,
producing [1m[37m62[0m.

[1m[37mWhat is the encryption weakness in your XMAS-encrypted list of numbers?[0m