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
#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

int dpd[4][2010];
inline int getint() {
	char c;
	int ret = 0;
	while ((c = getchar()) >= '0') ret = ret * 10 + c - '0';
	return ret;
}

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


	int N = getint();
	const int M = getint();

	dpd[0][0] = 1;
	auto dp0 = dpd[0];
	auto dp1 = dpd[1];
	auto dp2 = dpd[2];
	auto dp3 = dpd[3];

	rep(i, 3 * N) {
		for (int j = min(N, (i + 1) >> 1); j >= 0; j--) {
			ll tmp = dp0[j];
			dp3[j + 1] = tmp * (i + 1) * (i + 2) % M;
			dp2[j + 1] = (dp2[j + 1] + tmp * (i + 1)) % M;
			dp1[j] += tmp;
			if (dp1[j] > M) dp1[j] -= M;
		}
		dp3[0] = 0;
		swap(dp0, dp1);
		swap(dp1, dp2);
		swap(dp2, dp3);
	}

	ll kotae = 0;
	rep(i, N + 1) kotae += dp0[i];
	co(kotae % M);

	Would you please return 0;
}