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;
#define MN 200005
#define LL long long
#define mp make_pair
#define fir first
#define sec second
#define pii pair<int,int>
inline void chkmax(int &a,int b){if(a<b)a=b;}
inline void chkmin(int &a,int b){if(a>b)a=b;}
inline int read(){
	   int a=0,fh=1;char c=getchar();
	   while(c>'9'||c<'0'){if(c=='-')fh=-1;c=getchar();}
	   while('0'<=c&&c<='9'){
		   a=a*10+c-48;
		   c=getchar();
	   }
	   return a*fh;
}
const int py=6005;
int f[6005][16005],n,Mod;
void add(int &a,int b){((a+=b)>=Mod)?(a-=Mod):0;}
int main(){
	n=read()*3;Mod=read();
	f[0][py]=1;
	for(int i=0;i<n;++i){
		for(int j=-i;j<=i;++j){
			add(f[i+1][j+py+1],f[i][j+py]);
			add(f[i+2][j+py-1],(LL)f[i][j+py]*(i+1)%Mod);
			add(f[i+3][j+py],(LL)f[i][j+py]*(i+1)*(i+2)%Mod);
		}
	}
	LL ans=0;
	for(int i=0;i<=n;++i)ans+=f[n][i+py];
	printf("%lld\n",ans%Mod);
	return 0;
}















