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
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#include <bits/stdc++.h>
using namespace std;
template<class t> inline t read(t &x){
	char c=getchar();bool f=0;x=0;
	while(!isdigit(c)) f|=c=='-',c=getchar();
	while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	if(f) x=-x;return x;
}
template<class t,class ...A> inline void read(t &x,A &...a){
	read(x);read(a...);
}
template<class t> inline void write(t x){
	if(x<0) putchar('-'),write(-x);
	else{if(x>9) write(x/10);putchar('0'+x%10);}
}

const int N=6005,T=6000;
int n,mod,f[N][4][N<<1];
long long ans;

signed main(){
	read(n,mod);n*=3;
	f[1][0][T]=1;
	for(int i=1;i<n;i++) for(int j=0;j<3;j++) for(int k=-i;k<=i;k++){
		if(j<=1) f[i+1][j+1][k+T]=(f[i+1][j+1][k+T]+1ll*f[i][j][k+T]*i%mod)%mod;
		if(j==0) f[i+1][0][k+1+T]=(f[i+1][0][k+1+T]+f[i][j][k+T])%mod;
		else if(j==1) f[i+1][0][k-1+T]=(f[i+1][0][k-1+T]+f[i][j][k+T])%mod;
		else f[i+1][0][k+T]=(f[i+1][0][k+T]+f[i][j][k+T])%mod;
	}
	for(int k=0;k<=n;k+=3) ans=(ans+f[n][0][k-1+T]+f[n][1][k+1+T]+f[n][2][k+T])%mod;
	write(ans);
}