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 maxn=2005;
int n,mod;
int dp[maxn*3][maxn];
void Add(int &a,int b){
	a+=b;
	if(a>=mod)a-=mod;
}

int main(){
	scanf("%d%d",&n,&mod);
	dp[0][0]=1;
	for(int i=1;i<=3*n;i++)for(int j=0;j<=n;j++){
		Add(dp[i][j],dp[i-1][j]);
		if(i>=2&&j)Add(dp[i][j],1LL*dp[i-2][j-1]*(i-1)%mod);
		if(i>=3&&j)Add(dp[i][j],1LL*dp[i-3][j-1]*(i-1)%mod*(i-2)%mod);
	}
	int ans=0;
	for(int j=0;j<=n;j++)Add(ans,dp[3*n][j]);
	printf("%d\n",ans);
	return 0;
}