D. Fixing a Binary String

You are given a binary string 𝑠 of length 𝑛, consisting of zeros and ones. You can perform the following operation exactly once:

1. Choose an integer 𝑝 (1≤𝑝≤𝑛).
2. Reverse the substring 𝑠1𝑠2…𝑠𝑝. After this step, the string 𝑠1𝑠2…𝑠𝑛 will become 𝑠𝑝𝑠𝑝−1…𝑠1𝑠𝑝+1𝑠𝑝+2…𝑠𝑛.
3. Then, perform a cyclic shift of the string 𝑠 to the left 𝑝 times. After this step, the initial string 𝑠1𝑠2…𝑠𝑛 will become 𝑠𝑝+1𝑠𝑝+2…𝑠𝑛𝑠𝑝𝑠𝑝−1…𝑠1.

For example, if you apply the operation to the string 110001100110 with 𝑝=3, after the second step, the string will become 011001100110, and after the third step, it will become 001100110011.

A string 𝑠 is called 𝑘-proper if two conditions are met:

• 𝑠1=𝑠2=…=𝑠𝑘;
• 𝑠𝑖+𝑘≠𝑠𝑖 for any 𝑖 (1≤𝑖≤𝑛−𝑘).

For example, with 𝑘=3, the strings 000, 111000111, and 111000 are 𝑘-proper, while the strings 000000, 001100, and 1110000 are not.

You are given an integer 𝑘, which is a divisor of 𝑛. Find an integer 𝑝 (1≤𝑝≤𝑛) such that after performing the operation, the string 𝑠 becomes 𝑘-proper, or determine that it is impossible. Note that if the string is initially 𝑘-proper, you still need to apply exactly one operation to it.