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 ld long double
#define db double
#define pint pair<int,int>
#define mk(x,y) make_pair(x,y)
#define fi first
#define se second
#define Rep(x,y,z) for(int x=y;x<=z;x++)
#define Red(x,y,z) for(int x=y;x>=z;x--)
using namespace std;
const int MAXN=2e3+5;
char buf[1<<12],*pp1=buf,*pp2=buf,nc;int ny;
inline char gc() {return pp1==pp2&&(pp2=(pp1=buf)+fread(buf,1,1<<12,stdin),pp1==pp2)?EOF:*pp1++;}
//inline char gc(){return getchar();}
inline int read(){
	int x=0;for(ny=1;nc=gc(),(nc<48||nc>57)&&nc!=EOF;)if(nc==45)ny=-1;if(nc<0)return nc;
	for(x=nc-48;nc=gc(),47<nc&&nc<58&&nc!=EOF;x=(x<<3)+(x<<1)+(nc^48));return x*ny;
}
int n,Mod,F[MAXN*3][MAXN*4],ans;
inline void Add(int&x,int y){x+=y,x>=Mod?x-=Mod:0;}
int main(){
//	freopen("std.in","r",stdin);
//	freopen("std.out","w",stdout);
	n=read(),Mod=read(),F[0][n+1]=1;int now=0;
	Rep(i,0,3*n)Rep(j,-n,n*3)if(F[i][j+(n+1)]){
		int d=j+(n+1),tmp=F[i][d];
		Add(F[i+1][d+1],tmp),Add(F[i+2][d-1],1ll*tmp*(i+1)%Mod),Add(F[i+3][d],1ll*tmp*(i+2)%Mod*(i+1)%Mod);
	}Rep(t,n+1,4*n+1)ans=(ans+F[3*n][t])%Mod;cout<<ans<<'\n';
	return 0;
}


