A15539. Doping

We call an array $a$ of length $n$ fancy if for each $1 < i \le n$ it holds that $a_i = a_{i-1} + 1$ .

Let's call $f(p)$ applied to a permutation $^\dagger$ of length $n$ as the minimum number of subarrays it can be partitioned such that each one of them is fancy. For example $f([1,2,3]) = 1$ , while $f([3,1,2]) = 2$ and $f([3,2,1]) = 3$ .

Given $n$ and a permutation $p$ of length $n$ , we define a permutation $p'$ of length $n$ to be $k$ -special if and only if:

- $p'$ is lexicographically smaller $^\ddagger$ than $p$ , and
- $f(p') = k$ .

Your task is to count for each $1 \le k \le n$ the number of $k$ -special permutations modulo $m$ .

$^\dagger$ A permutation 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).

$^\ddagger$ A permutation $a$ of length $n$ is lexicographically smaller than a permutation $b$ of length $n$ if and only if the following holds: in the first position where $a$ and $b$ differ, the permutation $a$ has a smaller element than the corresponding element in $b$ .

The first line contains two integers $n$ and $m$ ( $1 \le n \le 2000$ , $10 \le m \le 10^9$ ) — the length of the permutation and the required modulo.

The second line contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ( $1 \le p_i \le n$ ) — the permutation $p$ .

Print $n$ integers, where the $k$ -th integer is the number of $k$ -special permutations modulo $m$ .

In the first example, the permutations that are lexicographically smaller than $[1,3,4,2]$ are:

- $[1,2,3,4]$ , $f([1,2,3,4])=1$ ;
- $[1,2,4,3]$ , $f([1,2,4,3])=3$ ;
- $[1,3,2,4]$ , $f([1,3,2,4])=4$ .

Thus our answer is $[1,0,1,1]$ .

In the second example, the permutations that are lexicographically smaller than $[3,2,1]$ are:

- $[1,2,3]$ , $f([1,2,3])=1$ ;
- $[1,3,2]$ , $f([1,3,2])=3$ ;
- $[2,1,3]$ , $f([2,1,3])=3$ ;
- $[2,3,1]$ , $f([2,3,1])=2$ ;
- $[3,1,2]$ , $f([3,1,2])=2$ .

Thus our answer is $[1,2,2]$ .