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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll mod;
ll dp[6666][2222];
int main()
{
	int num;
	scanf("%d%lld", &num, &mod);
	dp[0][0] = 1;
	for (int i = 0; i < num * 3; i++)
	{
		int t = num * 3 - i;
		for (int j = 0; j <= num; j++)
		{
			dp[i + 3][j + 1] = (dp[i + 3][j + 1] + dp[i][j] * (t - 1)*(t - 2)) % mod;
			dp[i + 2][j + 1] = (dp[i + 2][j + 1] + dp[i][j] * (t - 1)) % mod;
			dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % mod;
		}
	}
	ll ans = 0;
	for (int i = 0; i <= num; i++)ans += dp[num * 3][i];
	printf("%lld\n", ans%mod);
}