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;
#define int long long
const int N=2000+5,base=2001*3;
int f[N*3][N*6],mod;
void add(int &x,int y){
	x=(x+y)%mod;
}
signed main(){
	int n;
	cin>>n>>mod;
	n*=3;
	f[0][base]=1;
	for(int i=0;i<n;++i)
		for(int j=-i;j<=i;++j)
			if(f[i][j+base]){
				add(f[i+1][j+1+base],f[i][j+base]);
				add(f[i+2][j-1+base],1ll*f[i][j+base]*(i+1)%mod);
				add(f[i+3][j+base],1ll*f[i][j+base]*(i+1)%mod*(i+2)%mod);
			}
	int ans=0;
	for(int j=0;j<=n;++j)
		add(ans,f[n][j+base]);
	cout<<ans;
	return 0;
}