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;
using ll = long long;

ll D[6060][9090];
ll n, m, ans;

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

	ll i, j;

	cin >> n >> m;

	D[0][3000] = 1;
	for(i = 1; i <= n * 3; i ++){
		for(j = 0; j <= 9000; j ++){
			if(j > 0) D[i][j] += D[i - 1][j - 1];
			if(i > 1) D[i][j] += (i - 1) * D[i - 2][j + 1];
			if(i > 2) D[i][j] += (i - 1) * (i - 2) * D[i - 3][j];
			D[i][j] %= m;
		}
	}

	for(j = 3000; j <= 9000; j ++){
		ans = (ans + D[n * 3][j]) % m;
	}
	
	cout << ans << "\n";

	return 0;
}