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Count ints with each two consecutive 1s (in a binary representation) separated by at least K 0s.

In the binary number "`100100010000`" there are at least two 0s between any two consecutive 1s.

In the binary number "`100010000100010`" there are at least three 0s between any two consecutive 1s.

A positive integer N is called *K*-*sparse* if there are at least K 0s between any two consecutive 1s in its binary representation.

For example, the binary number "`100100010000`" is 2-sparse. Similarly, "`100010000100010`" is 3-sparse. It is also 2-sparse, because 2 < 3. It is also 1-sparse and 0-sparse.

We assume that any power of 2 (i.e. "`1`", "`10`", "`100`", "`1000`", ...) is K-sparse for any K.

Write a function:

class Solution { public int solution(String S, String T, int K); }

that, given:

- string S containing a binary representation of some positive integer A,
- string T containing a binary representation of some positive integer B,
- a positive integer K.

returns the number of K-sparse integers within the range [A..B] (both ends included). If the result exceeds 1,000,000,006, the function should return the remainder from the division of the result by 1,000,000,007.

For example, given S = "101" (A = 5), T = "1111" (B=15) and K=2, the function should return 2, because there are just two 2-sparse integers in the range [5..15], namely "`1000`" (i.e. 8) and "`1001`" (i.e. 9).

Write an ** efficient** algorithm for the following assumptions:

- K is an integer within the range [1..30];
- the length of S is within the range [1..300,000];
- the length of T is within the range [1..300,000];
- string S consists only of the characters "
0" and/or "1";- string T consists only of the characters "
0" and/or "1";- S and T have no leading zeros;
- A ≤ B.

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