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>
using namespace std;
template<typename T>inline T read(){
	T f=0,x=0;char c=getchar();
	while(!isdigit(c)) f=c=='-',c=getchar();
	while(isdigit(c)) x=x*10+c-48,c=getchar();
	return f?-x:x;
}
namespace run{
	const int N=6009;int mod;
	inline int add(int x,int y){return x+y>=mod?x-mod+y:x+y;}
	int f[N][(N<<1)+3],n;
	int main(){
		n=read<int>()*3,mod=read<int>();
		f[0][N]=1;
		for(int i=0;i<n;i++)
			for(int j=-i;j<=i;j++){
				f[i+1][j+1+N]=add(f[i+1][j+1+N],f[i][j+N]);
				f[i+2][j-1+N]=add(f[i+2][j-1+N],1LL*f[i][j+N]*(i+1)%mod);
				f[i+3][j+N]=add(f[i+3][j+N],1LL*f[i][j+N]*(i+1)%mod*(i+2)%mod);
			}

		int ans=0;
		for(int j=0;j<=n;j++) ans=add(ans,f[n][j+N]);
		printf("%d\n",ans);
		return 0;
	}
}
int main(){
#ifdef my
	freopen(".in","r",stdin);
	freopen(".out","w",stdout);
#endif
	return run::main();
}