Given a Binary Tree having positive and negative nodes, the task is to find maximum sum level in it.

Examples:

```
Input : 4
/ \
2 -5
/ \ /\
-1 3 -2 6
Output: 6
Explanation :
Sum of all nodes of 0'th level is 4
Sum of all nodes of 1'th level is -3
Sum of all nodes of 0'th level is 6
Hence maximum sum is 6
Input : 1
/ \
2 3
/ \ \
4 5 8
/ \
6 7
Output : 17
```

**Input:**

The task is to complete the method which takes one argument, root of Binary Tree. The Node has a data part which stores the data, pointer to left child and pointer to right child. There are multiple test cases. For each test case, this method will be called individually.

**Output:**

The function should return max sum level in the tree.

**Constraints:**

1 <=T<= 30

1 <=Number of nodes<= 100

-1000 <=Data of a node<= 1000

**Example(To be used only for expected output):
Input:**

2

2

1 2 L 1 3 R

6

4 2 L 4 -5 R 2 -1 L 2 3 R -5 -2 L -5 6 R

5

6

There are two test cases. First case represents a tree with 3 nodes and 2 edges where root is 1, left child of 1 is 2 and right child of 1 is 3. Second test case represents a tree with 6 edges and 7 nodes.

Author: Shubham Joshi 1

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