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>

const int N = 6005, M = N << 1, V = 6000;

int n, djq, f[N][M], ans;

int main()
{
	std::cin >> n >> djq; n *= 3;
	f[0][V] = 1;
	for (int i = 1; i <= n; i++)
		for (int j = -i + V; j <= i + V; j++)
		{
			if (i >= 3) f[i][j] = (1ll * (i - 1) * (i - 2)
				* f[i - 3][j] + f[i][j]) % djq;
			if (i >= 2) f[i][j] = (1ll * (i - 1) * f[i - 2][j + 1] + f[i][j]) % djq;
			f[i][j] = (f[i - 1][j - 1] + f[i][j]) % djq;
		}
	for (int i = V; i <= n + V; i++) ans = (ans + f[n][i]) % djq;
	return std::cout << ans << std::endl, 0;
}