You are given a binary string s that contains at least one '1'.You have to rearrange the bits in such a way that the resulting binary number is the maximum odd binary number that can be created from this combination.Return a string representing the maximum odd binary number that can be created from the given combination.Note that the resulting string can have leading zeros.

You are given a positive integer n.A binary string x is valid if all substrings of x of length 2 contain at least one "1".Return all valid strings with length n, in any order. Example 1:Input: n = 3Output: ["010","011","101","110","111"]Explanation:The valid strings of length 3 are: "010", "011", "101", "110", and "111".Example 2:Input: n = 1Output: ["0","1"]Explanation:The valid strings of length 1 are: "0" and "1".

Write a program that takes in a positive integer as input, and outputs a string of 1's and 0's representing the integer in reverse binary. For an integer x, the algorithm is:As long as x is greater than 0 Output x % 2 (remainder is either 0 or 1) x = x / 2Note: The above algorithm outputs the 0's and 1's in reverse order.Ex: If the input is:6the output is:0116 in binary is 110; the algorithm outputs the bits in reverse.

18. The number of 1’s in the binary string is evenGroup of answer choicesParityOdd parityEven parityBinary sequence

You are given a binary string s𝑠 of length n𝑛, consisting of zeros and ones. You can perform the following operation exactly once:Choose an integer p𝑝 (1≤p≤n1≤𝑝≤𝑛).Reverse the substring s1s2…sp𝑠1𝑠2…𝑠𝑝. After this step, the string s1s2…sn𝑠1𝑠2…𝑠𝑛 will become spsp−1…s1sp+1sp+2…sn𝑠𝑝𝑠𝑝−1…𝑠1𝑠𝑝+1𝑠𝑝+2…𝑠𝑛.Then, perform a cyclic shift of the string s𝑠 to the left p𝑝 times. After this step, the initial string s1s2…sn𝑠1𝑠2…𝑠𝑛 will become sp+1sp+2…snspsp−1…s1𝑠𝑝+1𝑠𝑝+2…𝑠𝑛𝑠𝑝𝑠𝑝−1…𝑠1.For example, if you apply the operation to the string 110001100110 with p=3𝑝=3, after the second step, the string will become 011001100110, and after the third step, it will become 001100110011.A string s𝑠 is called k𝑘-proper if two conditions are met:s1=s2=…=sk𝑠1=𝑠2=…=𝑠𝑘;si+k≠si𝑠𝑖+𝑘≠𝑠𝑖 for any i𝑖 (1≤i≤n−k1≤𝑖≤𝑛−𝑘).For example, with k=3𝑘=3, the strings 000, 111000111, and 111000 are k𝑘-proper, while the strings 000000, 001100, and 1110000 are not.You are given an integer k𝑘, which is a divisor of n𝑛. Find an integer p𝑝 (1≤p≤n1≤𝑝≤𝑛) such that after performing the operation, the string s𝑠 becomes k𝑘-proper, or determine that it is impossible. Note that if the string is initially k𝑘-proper, you still need to apply exactly one operation to it.InputEach test consists of multiple test cases. The first line contains one integer t𝑡 (1≤t≤1041≤𝑡≤104) — the number of test cases. The description of the test cases follows.The first line of each test case contains two integers n𝑛 and k𝑘 (1≤k≤n1≤𝑘≤𝑛, 2≤n≤1052≤𝑛≤105) — the length of the string s𝑠 and the value of k𝑘. It is guaranteed that k𝑘 is a divisor of n𝑛.The second line of each test case contains a binary string s𝑠 of length n𝑛, consisting of the characters 0 and 1.It is guaranteed that the sum of n𝑛 over all test cases does not exceed 2⋅1052⋅105.OutputFor each test case, output a single integer — the value of p𝑝 to make the string k𝑘-proper, or −1−1 if it is impossible.If there are multiple solutions, output any of them.ExampleinputCopy78 4111000014 2111012 31110001000115 5000006 11010018 40111000112 2110001100110outputCopy3-1754-13NoteIn the first test case, if you apply the operation with p=3𝑝=3, after the second step of the operation, the string becomes 11100001, and after the third step, it becomes 00001111. This string is 44-proper.In the second test case, it can be shown that there is no operation after which the string becomes 22-proper.In the third test case, if you apply the operation with p=7𝑝=7, after the second step of the operation, the string becomes 100011100011, and after the third step, it becomes 000111000111. This string is 33-proper.In the fourth test case, after the operation with any p𝑝, the string becomes 55-proper.

Given the binary representation of an integer as a string s, return the number of steps to reduce it to 1 under the following rules:If the current number is even, you have to divide it by 2.If the current number is odd, you have to add 1 to it.It is guaranteed that you can always reach one for all test cases.