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.

Examples

Input

1 998244353


Output

6


Input

2 998244353


Output

261


Input

314 1000000007


Output

182908545
#include<bits/stdc++.h>
#define ll long long
#define re register
#define cs const

using std::cerr;
using std::cout;

int mod=998244353;
inline int add(int a,int b){return a+b>=mod?a+b-mod:a+b;}
inline int dec(int a,int b){return a-b<0?a-b+mod:a-b;}
inline int mul(int a,int b){ll r=(ll)a*b;return r>=mod?r%mod:r;}
inline void Inc(int &a,int b){a+=b-mod;a+=a>>31&mod;}
inline void Dec(int &a,int b){a-=b;a+=a>>31&mod;}
inline void Mul(int &a,int b){a=mul(a,b);}

cs int N=6e3+7;

int n;
int f[N][N+N];

void Main(){
	scanf("%d%d",&n,&mod);
	f[0][N]=1;int ans=0;
	for(int re i=1;i<=3*n;++i){
		int l=N-i+1,r=N+i-1;
		for(int re j=l;j<=r;++j){
			if(i>0)Inc(f[i][j+1],f[i-1][j]);
			if(i>1)Inc(f[i][j-1],mul(f[i-2][j],i-1));
			if(i>2)Inc(f[i][j],mul(f[i-3][j],(i-1)*(i-2)));
		}
	}for(int re i=N;i<=N+3*n;++i)
		Inc(ans,f[3*n][i]);
	cout<<ans<<"\n";
}

inline void file(){
#ifdef zxyoi
	freopen("D.in","r",stdin);
#endif
}signed main(){file();Main();return 0;}