I like short descriptions and I think you so too.
Let’s say you have a number, N. Now try to find the last four digits of N!.
Actually It’s too easy.
I think you know that
$ N! = (1 \times 2 \times 3 \times ... \times N) $.
The input begins with a single integer indicating the number of test cases T (1 ≤ T ≤ 100). Each of the following test cases consists of a number N (0 ≤ N ≤ 1018).
For each test case output the last four digits of N factorial (N!).
If N! is less than 4 digits don’t forget to add 0s to left.
4 1 2 3 7
0001 0002 0006 5040
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