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;
const int N=6060;
int f[N][N*2],mod,n,m,ans;
int main(){
	scanf("%d%d",&n,&mod);m=n;n*=3;f[n][N]=1;
	for(int i=n;i;i--){
		for(int j=-(n-i);j<=(n-i);j++){
			int x=max(-j,0),y=max(j,0);
			int z=i-x*2-y;
			if(i>=3){
				if(z>0)f[i-3][j+N]=(f[i-3][j+N]+1ll*f[i][j+N]*(i-1)%mod*(i-2))%mod;
				else if(x>=2)f[i-3][j+3+N]=(f[i-3][j+3+N]+1ll*f[i][j+N]*(i-1)%mod*(i-2))%mod;
			}
			if(i>=2)f[i-2][j+1+N]=(f[i-2][j+1+N]+1ll*f[i][j+N]*(i-1))%mod;
			f[i-1][j-1+N]=(f[i-1][j-1+N]+f[i][j+N])%mod;
		}
	}
	for(int i=-n;i<=0;i++)if(i%3==0)(ans+=f[0][i+N])%=mod;
	cout<<ans;
	return 0;
}