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;

typedef long long LL;
#define N 6200

LL n,mod,f[N][N],g[N][N],ans;

void upd(LL &x,LL y){x=(x+y)%mod;}

LL C2(LL x){return x*(x-1)%mod;}

int main(){
	scanf("%lld%lld",&n,&mod);
	f[0][0]=1;;
	for (LL i=0;i<=n*3;++i){
		upd(f[i][0],g[i][0]);
		g[i][0]=f[i][0];
		for (LL j=0;j<=n*3;++j){
			upd(f[i+3][j],C2(n*3-i-1)*f[i][j]);
			upd(f[i+2][j+1],f[i][j]*(n*3-i-1));
			if (j) upd(f[i+1][j-1],f[i][j]);
			if (j) upd(g[i+3][j],C2(n*3-i-1)*g[i][j]);
			if (j) upd(g[i+2][j-1],g[i][j]*(n*3-i-1));
			upd(g[i+1][j+1],g[i][j]);
		}
	}
	for (LL i=0;i<=n*3;++i) upd(ans,g[n*3][i]);
	printf("%lld\n",ans);
	
	return 0;
}
			