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 cs const
using namespace std;
int Mod;
int add(int a, int b){ return a + b >= Mod ? a + b - Mod : a + b; }
int dec(int a, int b){ return a - b < 0 ? a - b + Mod : a - b; }
int mul(int a, int b){ return 1ll * a * b % Mod; }
void Add(int &a, int b){ a = add(a,b); }
void Dec(int &a, int b){ a = dec(a,b); }
void Mul(int &a, int b){ a = mul(a,b); }
cs int N = 6e3 + 50;
int n, f[N][N*2];
int main(){
	#ifdef FSYolanda
	freopen("1.in","r",stdin);
	#endif
	scanf("%d%d",&n,&Mod); n*=3; f[0][0+n]=1;
	for(int i=1,u=n; i<=u; i++){
		for(int j=-(i>>1); j<=i; j++){
			Add(f[i][j+1+n],f[i-1][j+n]);
			if(i>1) Add(f[i][j-1+n],mul(f[i-2][j+n],i-1));
			if(i>2) Add(f[i][j+n],mul(f[i-3][j+n],(i-1)*(i-2)));
		}
	} 
	int as=0;
	for(int i=n; i<=n+n; i++) Add(as,f[n][i]);
	cout<<as; return 0;
}