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 ll long long
#define ar array

const int mxN=2e3;
int n, m;
ll iv[3*mxN+1], f2[3*mxN+1], ans;

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);

	cin >> n >> m;
	iv[1]=1;
	for(int i=2; i<=3*mxN; ++i)
		iv[i]=m-m/i*iv[m%i]%m;
	f2[0]=1;
	for(int i=1; i<=3*mxN; ++i)
		f2[i]=f2[i-1]*iv[i]%m;
	ll p1=1;
	for(int x=0; x<=n; ++x, p1=p1*iv[3]%m) {
		ll p2=1;
		for(int y=0; x+y<=n; ++y, p2=p2*iv[2]%m) {
			int z=n-x-y;
			ans=(ans+p1*p2%m*f2[y]%m*f2[y+3*z]%m*f2[x])%m;
			//cout << x << " " << y << " " << z << endl;
			//cout << p1*p2%m*720%m*f2[y]%m*f2[y+3*z]%m*f2[x]%m << endl;
		}
	}
	for(int i=2; i<=3*n; ++i)
		ans=ans*i%m;
	cout << ans;
}