A little help/hint would be appreciated. Also, the teaser round had ended at 7 PM if you are wondering that I’m cheating or not.

Input : Standard Input

Output : Standard Output

Time Limit : 1 sec

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