Problem Statement   There is a narrow passage.
Inside the passage there are some wolves.
You are given a int[] size that contains the sizes of those wolves, from left to right.
The passage is so narrow that some pairs of wolves cannot pass by each other.
More precisely, two adjacent wolves may swap places if and only if the sum of their sizes is maxSizeSum or less.
Assuming that no wolves leave the passage, what is the number of different permutations of wolves in the passage?
Note that two wolves are considered different even if they have the same size.
Let X be the number of permutations of wolves that can be obtained from their initial order by swapping a pair of wolves zero or more times.
Compute and return the value (X modulo 1,000,000,007).   Definition   Class:  NarrowPassage2  Method:  count  Parameters:  int[], int  Returns:  int  Method signature:  int count(int[] size, int maxSizeSum)  (be sure your method is public) 
    Constraints    size will contain between 1 and 50 elements, inclusive.    Each element in size will be between 1 and 1,000,000,000, inclusive.    maxSizeSum will be between 1 and 1,000,000,000, inclusive.   Examples  0)     Returns: 2  From {1, 2, 3}, you can swap 1 and 2 to get {2, 1, 3}. But you can't get other permutations. 

 1)     Returns: 6  Here you can swap any two adjacent wolves. Thus, all 3! = 6 permutations are possible. 

 2)     Returns: 3  You can get {1, 2, 3}, {2, 1, 3} and {2, 3, 1}. 

 3)    {1,1,1,1,1,1,1,1,1,1,1,1,1}  2 
 Returns: 227020758  All of these wolves are different, even though their sizes are the same. Thus, there are 13! different permutations possible. 

 4)     5)    
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