Codeforces Round 842 (Div. 2) |
---|
Finished |
Consider a permutation$$$^\dagger$$$ $$$p$$$ of length $$$3n$$$. Each time you can do one of the following operations:
We can show that every permutation can be made sorted in increasing order using only these operations. Let's call $$$f(p)$$$ the minimum number of these operations needed to make the permutation $$$p$$$ sorted in increasing order.
Given $$$n$$$, find the sum of $$$f(p)$$$ over all $$$(3n)!$$$ permutations $$$p$$$ of size $$$3n$$$.
Since the answer could be very large, output it modulo a prime $$$M$$$.
$$$^\dagger$$$ A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).
The only line of input contains two numbers $$$n$$$ and $$$M$$$ ($$$1 \leq n \leq 10^6$$$, $$$10^8 \leq M \leq 10^9$$$). It is guaranteed that $$$M$$$ is a prime number.
Output the answer modulo $$$M$$$.
1 100009067
9
2 100000357
1689
69 999900997
193862705
In the first test case, all the permutations are:
Therefore, the answer is $$$0+1+1+2+2+3=9$$$.
Name |
---|