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>
#include<algorithm>
#include<cctype>
using namespace std;
#define G getchar()
int read()
{
	int x=0; char ch=G;
	for (;!isdigit(ch);ch=G);
	for (;isdigit(ch);ch=G) x=(x<<3)+(x<<1)+(ch^48);
	return x;
}
#undef G
typedef long long ll;
int Mod;
int fac[6010],inv[6010];

int n;
int dp[6010][9010],B=3000;
inline void add(int &x,ll y){x=(x+y)%Mod;}
int main()
{
	n=read()*3,Mod=read();
	fac[0]=1; for (int i=1;i<=n;i++) fac[i]=1LL*fac[i-1]*i%Mod;
	inv[1]=1; for (int i=2;i<=n;i++) inv[i]=1LL*(-Mod/i)*inv[Mod%i]%Mod;
	
	dp[0][B]=fac[n];
	for (int i=0;i<n;i++){
		for (int j=max(-(i/2),i-n)+B;j<=i+B;j++){
			if (!dp[i][j]) continue;
			add(dp[i+1][j+1],1LL*dp[i][j]*inv[i+1]);
			if (i+2<=n&&j-1>=0) add(dp[i+2][j-1],1LL*dp[i][j]*inv[i+2]);
			if (i+3<=n) add(dp[i+3][j],1LL*dp[i][j]*inv[i+3]);
		}
	}
	int ans=0;
	for (int i=B;i<=n+B;i++) ans=(ans+dp[n][i])%Mod;
	printf("%d\n",(ans+Mod)%Mod);
	return 0;
}