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>
#define ll long long
#define db long double
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define all(a) a.begin(), a.end()

using namespace std;

const int N = 6007;

int dp[N][2*N+1];

int main(){
#ifdef LOCAL
	freopen("A_input.txt", "r", stdin);
	//freopen("A_output.txt", "w", stdout);
#endif
	ios_base::sync_with_stdio(0);
	cin.tie(0);

	int n, mod;
	cin >> n >> mod;

	dp[0][N] = 1;
	dp[1][N+1] = 1;
	dp[2][N+2] = 1;
	dp[2][N-1] = 1;

	for (int i = 3; i <= 3*n; ++i) {
		for (int take = 0; take <= 2; ++take) {
			int cost = 0;
			if (take == 0) cost = 1;
			if (take == 1) cost = -1;

			for (int was = 0; was < 2*N+1; ++was) {
				if (!dp[i-take-1][was]) continue;
				int ways = 1;
				if (take == 1) {
					ways = i-1;
				}
				if (take == 2) {
					ways = ((ll) (i-1) * (i-2)) % mod;
				}
				dp[i][was + cost] += (((ll) dp[i-take-1][was] * ways) % mod);
				if (dp[i][was + cost] >= mod) dp[i][was + cost] -= mod;
			}
		}
	}

	int ans = 0;
	for (int i = N; i < 2*N+1; ++i) {
		ans += dp[3*n][i];
		ans %= mod;
	}

	cout << ans;


}
