Solve the programming task below in a Python markdown code block.
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.

Given a binary string $S$ consisting of $0's$ and $1's$, find whether there exists a rightwise  [circular rotation] of the string such that every 2 adjacent $1's$ are separated by at most $C$ $0's$.

Note: The last occurrence of $1$ in the rotated string won't have any $1$ next to it, i.e, first and the last ones in the string are not considered to be adjacent.

------ Input: ------

First line will contain $T$, number of testcases. Then the testcases follow. 
Each testcase contains of two lines of input.
First line contains two space separated integers $N, C$, length of the string and the upper limit on a number of $0's$ between $2$ adjacent $1's$.
Second line contains the binary string $S$.

------ Output: ------
For each testcase, output in a single line "YES" if there exists a rightwise circular rotation of string satisfying the criteria and "NO" if it doesn't exist.

Note: The output is case-insensitive ― each letter may be printed in upper case or lower case.

------ Constraints  ------
$1 ≤ N ≤ 5*10^{5}$
$0 ≤ C ≤ max(0, N - 2)$
Sum of $N$ over all tests is atmost $10^{6}$.

----- Sample Input 1 ------ 
3

4 1

1100

4 0

0101

6 1

101001
----- Sample Output 1 ------ 
YES

NO

YES
----- explanation 1 ------ 
Case 1: In the original configuration maximum number of $0's$ between $2$ adjacent $1's$ is $0$, therefore it satisfies the criteria.

Case 2: The $4$ circular rotations of the string S = {"0101", "1010", "0101", "1010"}. In all the cases the maximum number of $0's$ between $2$ consecutive $1's$ is $1$ which doesn't satisfies the criteria.

Case 3: The $6$ circular rotations of the string S = {"101001", "110100", "011010", "001101", "100110", "010011"} out of which $second$, $third$ and $fourth$ strings satisfy the criteria.