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 <iostream>
#include <vector>
#include <string>
using namespace std;

using cat = long long;

cat pw(cat a, cat e, cat mod) {
	if(e <= 0) return 1;
	cat x = pw(a, e/2, mod);
	x = x * x % mod;
	return e&1 ? x * a % mod : x;
}

int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int N, M;
	cin >> N >> M;
	vector<cat> invf(3*N+1, 1), inv2(N+1, 1), inv3(N+1, 1);
	for(int i = 1; i <= 3*N; i++) invf[i] = invf[i-1] * pw(i, M-2, M) % M;
	for(int i = 1; i <= N; i++) inv2[i] = inv2[i-1] * pw(2, M-2, M) % M;
	for(int i = 1; i <= N; i++) inv3[i] = inv3[i-1] * pw(3, M-2, M) % M;
	cat ans = 0;
	for(int k3 = 0; k3 <= N; k3++) for(int k2 = 0; k2 <= N-k3; k2++) {
		int k1 = 3*N-2*k2-3*k3;
		ans += inv2[k2] * inv3[k3] % M * invf[k1] % M * invf[k2] % M * invf[k3] % M;
		if(ans >= M) ans -= M;
	}
	for(int i = 1; i <= 3*N; i++) ans = ans * i % M;
	cout << ans << "\n";
}
