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<cstdio>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<cstring>
using namespace std;
inline int read()
{
	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 stack[20];
inline void write(int x)
{
	if(x<0){putchar('-');x=-x;}
	if(!x){putchar('0');return;}
	int top=0;
	while(x)stack[++top]=x%10,x/=10;
	while(top)putchar(stack[top--]+'0');
}
inline void pr1(int x){write(x),putchar(' ');}
inline void pr2(int x){write(x),puts("");}
int mod;
inline int ad(int x){return x>=mod?x-mod:x;}
int f[6010][2010];
int main()
{
    //freopen("a.in","r",stdin);
    //freopen("a.out","w",stdout);
    int n=read(),ans=0;mod=read();
    f[0][0]=1;
    for(int i=0;i<n*3;i++)
    {
    	for(int j=0;j<=n;j++)
    	{
	    	for(int k=1;k<=3;k++)
	    	{
	    		int uf=j+1,ul=1;
	    		if(k==1)uf--;
	    		else
				{
					ul=1LL*ul*(i+k-1)%mod;
					if(k==3)ul=1LL*ul*(i+k-2)%mod;
				}
	    		if(uf>n)continue;
	    		f[i+k][uf]=ad(f[i+k][uf]+1LL*f[i][j]*ul%mod);
	    	}
	    }
    }
	for(int i=0;i<=n;i++)ans=ad(ans+f[n*3][i]);
	pr2(ans);
    return 0;
}