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>
#define int long long
int n,mod,ans;
int f[6010][8010];
signed main(){
	scanf("%lld%lld",&n,&mod);
	f[0][n+1]=1;
	for(int i=0;i<=3*n;i++)
		for(int j=-n;j<=3*n;j++){
			(f[i+1][j+1+n+1]+=f[i][j+n+1])%=mod;
			(f[i+2][j-1+n+1]+=f[i][j+n+1]*(i+1)%mod)%=mod;
			(f[i+3][j+n+1]+=f[i][j+n+1]*(i+1)%mod*(i+2)%mod)%=mod;	
		}
	for(int i=0;i<=n;i++)
		(ans+=f[3*n][3*i+n+1])%=mod;
	printf("%lld\n",ans);
}