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>
//#include <ext/pb_ds/assoc_container.hpp>
 
//using namespace __gnu_pbds;
 
#define N 2000001
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
 
const ll INF = 1e18, MOD = 1e9 + 7, MOD2 = 1e6 + 3;

ll d[10001][10003], n, P, offset = 3000;

int main () {
	cin >> n >> P;

	d[0][offset] = 1;
	d[1][offset + 1] = 1;

	for (int i = 2; i <= 3 * n; i++)
		for (int j = - i / 2; j <= i; j++) {
			d[i][offset + j] += d[i-1][offset + j - 1];
			if (d[i][offset + j] >= P) d[i][offset + j] -= P;
			d[i][offset + j] += d[i-2][offset + j + 1] * (i - 1) % P;
			if (d[i][offset + j] >= P) d[i][offset + j] -= P;
			if (i > 2) d[i][offset + j] += d[i-3][offset + j] * (i - 1) * (i - 2) % P;
		}

	ll ans = 0;

	for (int i = 0; i <= 3 * n; i++) {
		ans += d[3 * n][offset + i];
		if (ans >= P) ans -= P;
	}

	cout << ans;
}