You are given a binary tree. Your task is to return the count of the diagonal paths to the leaf of the given binary tree such that all the values of the nodes on the diagonal are equal.

Note:

A diagonal traversal Consider lines of slope -1 passing between nodes.
         1
       /    \
     4        2
    /  \        \
   8    5         3
       /  \       /
      9    7      6

Example: Here potential diagonals are: 
1 -- 2 -- 3
4 -- 5 -- 7 -- 6
8 -- 9 -- 7 

For Example

Input: 

     5
  /  \
 6      5
  \       \
   6       5

Output: 2 

Explanation: 

Diagonal 6 – 6 and 5 – 5 contains equal value. Therefore, the required output is 2.

Input Format:

The first line of input contains an integer ‘T’ denoting the number of test cases.
The next ‘T’ lines represent the ‘T’ test cases.

The first line of input contains the elements of the tree in the level order form separated by a single space.
If any node does not have a left or right child, take -1 in its place. Refer to the example below.

Example:

Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted below :

        1
       /    \
     4        2
    /  \        \
   8    5         3
       /   \     /
       9    7   6


 1
 4 2
 8 5 -1 3
-1 -1 9 7 6 -1
-1 -1 -1 -1 -1 -1

Explanation :
Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 4
Right child of 1 = 2

Level 3 :
Left child of 4 = 8
Right child of 4 = 5
Left child of 2 = null (-1)
Right child of 2 = 3

Level 4 :
Left child of 8 = null (-1)
Right child of 8 = null (-1)
Left child of 5 = 9
Right child of 5 = 7
Left child of3 = 6
Right child of 5 = null (-1)

Level 5 :
Left child of 9 = null (-1)
Right child of 9 = null (-1)
Left child of 7 = null (-1)
Right child of 7 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)



The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not- 
null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null (-1).

Note :

The above format was just to provide clarity on how the input is formed for a given tree. 

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 4 2 8 5 -1 3 -1 -1 9 7 6 -1 -1 -1 -1 -1 -1 -1

Output Format:

For every test case, return the count of the diagonal paths to the leaf of the given binary tree such that all the values of the nodes on the diagonal are equal.

Constraint :

1 <= T <= 100
1 <= N <= 3000
1 <= data <= 10^9

Where ‘T’ represents the number of test cases, ‘N’ is the number of nodes in the tree, and data denotes data contained in the node of a binary tree.

Time Limit: 1 sec