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
//Created Time:2020年03月22日 星期日 20时16分29秒
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 6005

using namespace std;

int n, mod;
int f[N][N << 1], inv[N];

int main(){
	cin >> n >> mod; n = n * 3;
	int fac = 1; inv[1] = 1;
	for(int i = 1; i <= n; ++i)
		fac = 1ll * fac * i % mod;
	for(int i = 2; i <= n; ++i)
		inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
	f[0][n] = 1;
	for(int i = 0; i <= n; ++i)
		for(int j = 0; j <= n * 2; ++j){
			int w = f[i][j]; if(!w) continue;
			(f[i + 1][j + 1] += 1ll * w * inv[i + 1] % mod) %= mod;
			(f[i + 2][j - 1] += 1ll * w * inv[i + 2] % mod) %= mod;
			(f[i + 3][j] += 1ll * w * inv[i + 3] % mod) %= mod;
		}
	int res = 0;
	for(int j = n; j <= n * 2; ++j) (res += f[n][j]) %= mod;
	res = 1ll * res * fac % mod;
	cout << res << endl;
	return 0;
}