Problem:
Score : 1200 points

Problem Statement

Given is a positive integer N.
Find the number of permutations (P_1,P_2,\cdots,P_{3N}) of (1,2,\cdots,3N) that can be generated through the procedure below.
This number can be enormous, so print it modulo a prime number M.

Make N sequences A_1,A_2,\cdots,A_N of length 3 each, using each of the integers 1 through 3N exactly once.

Let P be an empty sequence, and do the following operation 3N times.

Among the elements that are at the beginning of one of the sequences A_i that is non-empty, let the smallest be x.

Remove x from the sequence, and add x at the end of P.

Constraints

1 \leq N \leq 2000

10^8 \leq M \leq 10^9+7

M is a prime number.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of permutations modulo M.

Sample Input 1

1 998244353

Sample Output 1

6

All permutations of length 3 count.

Sample Input 2

2 998244353

Sample Output 2

261

Sample Input 3

314 1000000007

Sample Output 3

182908545

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N; ll M; cin >> N >> M;
    vector<vector<ll>> dp(3*N+1,vector<ll>(N+1,0));
    dp[0][0]=1;
    for (int i=0;i<3*N;++i){
        for (int j=0;j<=N;++j){
            (dp[i+1][j]+=dp[i][j])%=M;
            if (i+2<=3*N&&j+1<=N) (dp[i+2][j+1]+=dp[i][j]*(i+1)%M)%=M;
            if (i+3<=3*N&&j+1<=N) (dp[i+3][j+1]+=dp[i][j]*(i+2)*(i+1)%M)%=M;
        }
    }
    ll ans=0;
    for (int j=0;j<=N;++j) (ans+=dp[3*N][j])%=M;
    cout << ans << '\n';
}