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>

#define pb push_back
#define mp make_pair
#define fi first
#define se second

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

template <typename T> bool chkmax(T &x,T y){return x<y?x=y,true:false;}
template <typename T> bool chkmin(T &x,T y){return x>y?x=y,true:false;}

int readint(){
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

int n,cys;
ll d[6005][2005],inv[6005];

ll mod(ll x){return x>=cys?x-cys:x;}

ll qpow(ll x,ll p){
	ll ret=1;
	for(;p;p>>=1,x=x*x%cys) if(p&1) ret=ret*x%cys;
	return ret;
}

int main(){
	n=readint()*3; cys=readint();
	inv[0]=1;
	for(int i=1;i<=n;i++) inv[i]=qpow(i,cys-2);
	d[0][0]=1;
	ll ans=0;
	for(int i=0;i<=n;i++){
		for(int j=0;j<=n/3;j++){
			d[i][j]=d[i][j]*inv[i]%cys;
			if(i==n) ans=mod(ans+d[i][j]);
			for(int k=i+1;k<=i+3&&k<=n;k++)
				d[k][j+(k-i>=2)]=mod(d[k][j+(k-i>=2)]+d[i][j]);
		}
	}
	for(int i=1;i<=n;i++) ans=ans*i%cys;
	printf("%lld\n",ans);
	return 0;
}