Count The Number

It is an interesting problem to count the number of unordered partitions of a number N.
This is generally done by putting N non-negative numbers in non-increasing order along a 1xN board, which sum to N.

In this problem however, you are given a 2xN board. You need to fill in non-negative numbers in this board in such a way, that
(1) The sum of all the numbers filled = N
(2) Each of the 2 rows consist of numbers in non-increasing order
(3) Each of the N columns consist of numbers in non-increasing order.

In how many ways can this be done, given the number N? Two ways are considered different if there is a cell in the board which has different numbers.
Input:

The first line consists of the number of testcases T. Each testcase consists of a single integer N.
Output:

For each testcase, output a single line, giving the number of ways.
Constraints:
T <= 10
N <= 70
The answer will fit in a 64 bit signed integer.
Sample Input:

1
5

Sample Output:
16

This may be helpful to you :slight_smile:

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