OI Problems

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## 题目描述

• $p$ 的前 $2n$ 个元素从小到大排序。
• $p$ 的后 $2n$ 个元素从小到大排序。

$f(p)$ 表示将 $p$ 从小到大排序所需的最小总操作次数。

## 输出格式

1 100009067

9

2 100000357

1689

69 999900997

193862705


## 提示

In the first test case, all the permutations are:

• $[1, 2, 3]$, which requires $0$ operations;
• $[1, 3, 2]$, which requires $1$ operation;
• $[2, 1, 3]$, which requires $1$ operation;
• $[2, 3, 1]$, which requires $2$ operations;
• $[3, 1, 2]$, which requires $2$ operations;
• $[3, 2, 1]$, which requires $3$ operations.

Therefore, the answer is $0+1+1+2+2+3=9$.

Consider a permutation $^\dagger p$ of length $3n$ . Each time you can do one of the following operations:

• Sort the first $2n$ elements in increasing order.
• Sort the last $2n$ elements in increasing order.

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).

## Input

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

Output the answer modulo $M$.

1 100009067

9

2 100000357

1689

69 999900997

193862705


## Tip

In the first test case, all the permutations are:

• $[1, 2, 3]$, which requires $0$ operations;
• $[1, 3, 2]$, which requires $1$ operation;
• $[2, 1, 3]$, which requires $1$ operation;
• $[2, 3, 1]$, which requires $2$ operations;
• $[3, 1, 2]$, which requires $2$ operations;
• $[3, 2, 1]$, which requires $3$ operations.

Therefore, the answer is $0+1+1+2+2+3=9$.